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Objectives
After completing this section, you should be able to
1. identify the ultraviolet region of the electromagnetic spectrum which is of most use to organic chemists.
2. interpret the ultraviolet spectrum of 1,3-butadiene in terms of the molecular orbitals involved.
3. describe in general terms how the ultraviolet spectrum of a compound differs from its infrared and NMR spectra.
Key Terms
Make certain that you can define, and use in context, the key term below.
• ultraviolet (UV) spectroscopy
• Molar absorptivity
Study Notes
Ultraviolet spectroscopy provides much less information about the structure of molecules than do the spectroscopic techniques studied earlier (infrared spectroscopy, mass spectroscopy, and NMR spectroscopy). Thus, your study of this technique will be restricted to a brief overview. You should, however, note that for an organic chemist, the most useful ultraviolet region of the electromagnetic spectrum is that in which the radiation has a wavelength of between 200 and 400 nm.
UV-Visible Absorption Spectra
To understand why some compounds are colored and others are not, and to determine the relationship of conjugation to color, we must make accurate measurements of light absorption at different wavelengths in and near the visible part of the spectrum. Commercial optical spectrometers enable such experiments to be conducted with ease, and usually survey both the near ultraviolet and visible portions of the spectrum. Ultraviolet-visible absorption spectroscopy provides much less information about the structure of molecules than do the spectroscopic techniques studied earlier (infrared spectroscopy, mass spectroscopy, and NMR spectroscopy) and mainly provides information about conjugated pi systems. For an organic chemist the most useful ultraviolet region of the electromagnetic spectrum involves radiation with a wavelength between 200 and 400 nm. UV/Vis absorption spectra also involve radiation from the visible region of the electromagnetic spectrum with wavelengths between 400 and 800 nm.
A diagram highlighting the various kinds of electronic excitation that may occur in organic molecules is shown below. Of the six transitions outlined, only the two lowest energy ones, n to pi* and pi to pi* (colored blue) are achieved by the energies available in the 200 to 800 nm range of a UV/VIs spectrum. These energies are sufficient to promote or excite a molecular electron to a higher energy orbital in many conjugated compounds.
When sample molecules are exposed to light having an energy that matches a possible electronic transition within the molecule, some of the light energy will be absorbed as the electron is promoted to a higher energy orbital. A UV/Vis spectrometer records the wavelengths at which absorption occurs, together with the degree of absorption at each wavelength. Absorbance, abbreviated 'A', is a unitless number which contains the same information as the 'percent transmittance' number used in IR spectroscopy. To calculate absorbance at a given wavelength, the computer in the spectrophotometer simply takes the intensity of light at that wavelength before it passes through the sample (I0), divides this value by the intensity of the same wavelength after it passes through the sample (I), then takes the log10 of that number:
$A = \log \left (\dfrac{I_0}{I} \right) \nonumber$
The resulting spectrum is presented as a graph of absorbance (A) versus wavelength, as in the isoprene spectrum shown below. Since isoprene is colorless, it does not absorb in the visible part of the spectrum and this region is not displayed on the graph. Notice that the convention in UV-vis spectroscopy is to show the baseline at the bottom of the graph with the peaks pointing up. Wavelength values on the x-axis are generally measured in nanometers (nm) rather than in cm-1 as is the convention in IR spectroscopy.
Typically, there are two things that are noted and recorded from a UV-Vis spectrum. The first is λmax, which is the wavelength at maximal light absorbance. As you can see, isoprene has λmax, = 222 nm. The second valuable piece of data is the absorbance at the λmax. In the isoprene spectrum the absorbance at the value λmax of 222 nm is about 0.8.
Molar absorptivity
Molar absorptivity ($epsilon$) is a physical constant, characteristic of the particular substance being observed and thus characteristic of the particular electron system in the molecule. Molar absorptivities may be very large for strongly absorbing chromophores (>100,000) and very small if absorption is weak (10 to 100). The magnitude of $epsilon$ reflects both the size of the chromophore and the probability that light of a given wavelength will be absorbed when it strikes the chromophore. Molar absorptivity ($\epsilon$) is defined via Beer's law as:
$\epsilon = \dfrac{A}{c\; l} \nonumber$
where
• $A$ is the sample absorbance
• $c$ is the sample concentration in moles/liter
• $l$ is the length of light path through the sample in cm
If the isoprene spectrum show above was obtained from a dilute hexane solution (c = 4 * 10-5 moles per liter) in a 1 cm sample cuvette, a simple calculation using the above formula indicates a molar absorptivity of 20,000 at the maximum absorption wavelength, symbolized as $λ_{max}$.
The only molecular moieties likely to absorb light in the 200 to 800 nm region are functional groups that contain pi-electrons and hetero atoms having non-bonding valence-shell electron pairs. Such light absorbing groups are referred to as chromophores. A list of some simple chromophores and their light absorption characteristics are provided below. The oxygen non-bonding electrons in alcohols and ethers do not give rise to absorption above 160 nm. Consequently, pure alcohol and ether solvents may be used for spectroscopic studies.
Chromophore Example Excitation λmax, nm ε @ λmax Solvent
C=C Ethene π → π* 171 15,000 hexane
C≡C 1-Hexyne π → π* 180 10,000 hexane
C=O Ethanal n → π*
π → π*
290
180
15
10,000
hexane
hexane
N=O Nitromethane
n → π*
π → π*
275
200
17
5,000
ethanol
ethanol
C-X X=Br
X=I
Methyl bromide
Methyl Iodide
n → σ*
n → σ*
205
255
200
360
hexane
hexane
Electronic Transitions (cause of UV-Visible absorption)
As previously noted, electronic transitions in organic molecules lead to UV and visible absorption. As a rule, energetically favored electron promotion will be from the highest occupied molecular orbital (HOMO) to the lowest unoccupied molecular orbital (LUMO), and the resulting species is called an excited state. The molecular orbital picture for the hydrogen molecule (H2) consists of one bonding σ MO, and a higher energy antibonding σ* MO. When the molecule is in the ground state, both electrons are paired in the lower-energy bonding orbital – this is the Highest Occupied Molecular Orbital (HOMO). The antibonding σ* orbital, in turn, is the Lowest Unoccupied Molecular Orbital (LUMO).
If the molecule is exposed to light of a wavelength with energy equal to ΔE, the HOMO-LUMO energy gap, this wavelength will be absorbed and the energy used to bump one of the electrons from the HOMO to the LUMO – in other words, from the σ to the σ* orbital. This is referred to as a σ - σ* transition. ΔE for this electronic transition is 258 kcal/mol, corresponding to light with a wavelength of 111 nm.
When a double-bonded molecule such as ethene (common name ethylene) absorbs light, it undergoes a π - π* transition. Because π- π* energy gaps are narrower than σ - σ* gaps, ethene absorbs light at 165 nm - a longer wavelength than molecular hydrogen.
The electronic transitions of both molecular hydrogen and ethene are too energetic to be accurately recorded. Where electronic transition becomes useful to most organic and biological chemists is in the study of molecules with conjugated pi systems. In these groups, the energy gap for π -π* transitions is smaller than for isolated double bonds, and thus the wavelength absorbed is longer. The MO diagram for 1,3-butadiene, the simplest conjugated system. Recall that we can draw a diagram showing the four pi MO’s that result from combining the four 2pz atomic orbitals. The lower two orbitals are pi bonding, while the upper two are pi antibonding.
Comparing this MO picture to that of ethene, our isolated pi-bond example, we see that the HOMO-LUMO energy gap is indeed smaller for the conjugated system. 1,3-butadiene absorbs UV light with a wavelength of 217 nm.
As conjugated pi systems become larger, the energy gap for a π - π* transition becomes increasingly narrow, and the wavelength of light absorbed correspondingly becomes longer. The absorbance due to the π - π* transition in 1,3,5-hexatriene, for example, occurs at 258 nm, corresponding to a ΔE of 111 kcal/mol.
In molecules with extended pi systems, the HOMO-LUMO energy gap becomes so small that absorption occurs in the visible rather then the UV region of the electromagnetic spectrum. Beta-carotene, with its system of 11 conjugated double bonds, absorbs light with wavelengths in the blue region of the visible spectrum while allowing other visible wavelengths – mainly those in the red-yellow region - to be transmitted. This is why carrots are orange.
The conjugated pi system in 4-methyl-3-penten-2-one gives rise to a strong UV absorbance at 236 nm due to a π - π* transition. However, this molecule also absorbs at 314 nm. This second absorbance is due to the transition of a non-bonding (lone pair) electron on the oxygen up to a π* antibonding MO:
This is referred to as an $n - π^*$ transition. The nonbonding ($n$) MO’s are higher in energy than the highest bonding p orbitals, so the energy gap for an $n - π^*$ transition is smaller that that of a $π - π^*$ transition – and thus the $n - π^*$ peak is at a longer wavelength. In general, $n - π^*$ transitions are weaker (less light absorbed) than those due to $π - π^*$ transitions.
Use of UV/Vis Spectroscopy in Biological Systems
The bases of DNA and RNA are good chromophores:
Biochemists and molecular biologists often determine the concentration of a DNA sample by assuming an average value of ε = 0.020 ng-1×mL for double-stranded DNA at its λmax of 260 nm (notice that concentration in this application is expressed in mass/volume rather than molarity: ng/mL is often a convenient unit for DNA concentration when doing molecular biology).
Because the extinction coefficient of double stranded DNA is slightly lower than that of single stranded DNA, we can use UV spectroscopy to monitor a process known as DNA melting. If a short stretch of double stranded DNA is gradually heated up, it will begin to ‘melt’, or break apart, as the temperature increases (recall that two strands of DNA are held together by a specific pattern of hydrogen bonds formed by ‘base-pairing’).
As melting proceeds, the absorbance value for the sample increases, eventually reaching a high plateau as all of the double-stranded DNA breaks apart, or ‘melts’. The mid-point of this process, called the ‘melting temperature’, provides a good indication of how tightly the two strands of DNA are able to bind to each other.
Later we will see how the Beer - Lambert Law and UV spectroscopy provides us with a convenient way to follow the progress of many different enzymatic redox (oxidation-reduction) reactions. In biochemistry, oxidation of an organic molecule often occurs concurrently with reduction of nicotinamide adenine dinucleotide (NAD+, the compound whose spectrum we saw earlier in this section) to NADH:
Both NAD+ and NADH absorb at 260 nm. However NADH, unlike NAD+, has a second absorbance band with λmax = 340 nm and ε = 6290 L*mol-1*cm-1. The figure below shows the spectra of both compounds superimposed, with the NADH spectrum offset slightly on the y-axis:
By monitoring the absorbance of a reaction mixture at 340 nm, we can 'watch' NADH being formed as the reaction proceeds, and calculate the rate of the reaction.
UV spectroscopy is also very useful in the study of proteins. Proteins absorb light in the UV range due to the presence of the aromatic amino acids tryptophan, phenylalanine, and tyrosine, all of which are chromophores.
Biochemists frequently use UV spectroscopy to study conformational changes in proteins - how they change shape in response to different conditions. When a protein undergoes a conformational shift (partial unfolding, for example), the resulting change in the environment around an aromatic amino acid chromophore can cause its UV spectrum to be altered.
Exercise $1$
1. 50 microliters of an aqueous sample of double stranded DNA is dissolved in 950 microliters of water. This diluted solution has a maximal absorbance of 0.326 at 260 nm. What is the concentration of the original (more concentrated) DNA sample, expressed in micrograms per microliter?
2. What is the energy range for 300 nm to 500 nm in the ultraviolet spectrum? How does this compare to energy values from NMR and IR spectroscopy?
3. Identify all isolated and conjugated pi bonds in lycopene, the red-colored compound in tomatoes. How many pi electrons are contained in the conjugated pi system?
Answer
1) Using ε = A/c, we plug in our values for ε and A and find that c = 3.27 x 10-5M, or 32.7 mM.
2)
E = hc/λ
E = (6.62 × 10−34 Js)(3.00 × 108 m/s)/(3.00 × 10−7 m)
E = 6.62 × 10−19 J
The range of 3.972 × 10-19 to 6.62 × 10-19 joules. This energy range is greater in energy than the in NMR and IR.
3)
Objective
After completing this section, you should be able to use data from ultraviolet spectra to assist in the elucidation of unknown molecular structures.
Study Notes
It is important that you recognize that the ultraviolet absorption maximum of a conjugated molecule is dependent upon the extent of conjugation in the molecule.
The Importance of Conjugation
A comparison of the UV/Vis absorption spectrum of 1-butene, λmax = 176 nm, with that of 1,3-butadiene, λmax = 292 nm, clearly demonstrates that the effect of increasing conjugation is to shift toward longer wavelength (lower frequency, lower energy) absorptions. Further evidence of this effect is shown below. The spectrum on the left illustrates that conjugation of double and triple bonds also shifts the absorption maximum to longer wavelengths. From the polyene spectra displayed in the right it is clear that each additional double bond in the conjugated pi-electron system increases the absorption maximum about 30 nm. Also, the molar absorptivity (ε) roughly doubles with each new conjugated double bond. Spectroscopists use the terms defined in the table below when describing shifts in absorption. Thus, extending conjugation generally results in bathochromic and hyperchromic shifts in absorption.
Terminology for Absorption Shifts
Nature of Shift Descriptive Term
To Longer Wavelength Bathochromic
To Shorter Wavelength Hypsochromic
To Greater Absorbance Hyperchromic
To Lower Absorbance Hypochromic
Many other kinds of conjugated pi-electron systems act as chromophores and absorb light in the 200 to 800 nm region. These include unsaturated aldehydes and ketones and aromatic ring compounds. A few examples are displayed below. The spectrum of the unsaturated ketone (on the left) illustrates the advantage of a logarithmic display of molar absorptivity. The $\pi \rightarrow \pi^*$ absorption located at 242 nm is very strong, with an ε = 18,000. The weak $n \rightarrow \pi^*$ absorption near 300 nm has an ε = 100.
Benzene exhibits very strong light absorption near 180 nm (ε > 65,000) , weaker absorption at 200 nm (ε = 8,000) and a group of much weaker bands at 254 nm (ε = 240). Only the last group of absorptions are completely displayed because of the 200 nm cut-off characteristic of most spectrophotometers. The added conjugation in naphthalene, anthracene and tetracene causes bathochromic shifts of these absorption bands, as displayed in the chart below. All the absorptions do not shift by the same amount, so for anthracene (green shaded box) and tetracene (blue shaded box) the weak absorption is obscured by stronger bands that have experienced a greater red shift. As might be expected from their spectra, naphthalene and anthracene are colorless (with their absorptions in the UV range), but tetracene is orange (since its absorptions move into the visible range).
Looking at UV-Vis Spectra
Below is the absorbance spectrum of an important biological molecule called nicotinamide adenine dinucleotide, abbreviated NAD+. This compound absorbs light in the UV range due to the presence of conjugated pi-bonding systems.
Below is the absorbance spectrum of the common food coloring Red #3. The extended system of conjugated pi bonds causes the molecule to absorb light in the visible range. Because the λmax of 524 nm falls within the green region of the spectrum, the compound appears red to our eyes (recalling the color wheel from Section 14.7).
Example 14.8.2
How large is the π - π* transition in 4-methyl-3-penten-2-one?
Solution
Example 14.8.3
Which of the following molecules would you expect absorb at a longer wavelength in the UV region of the electromagnetic spectrum? Explain your answer.
Solution
Exercise $1$
Which of the following would show UV absorptions in the 200-300 nm range?
Answer
B and D would be in that range. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/14%3A_Delocalized_Pi_Systems%3A_Investigation_by_Ultraviolet_and_Visible_Spectroscopy/14.11%3A_%09Electronic_Spectra%3A__Ultraviolet_and__Visible__Spectrosc.txt |
Objectives
After completing this section, you should be able to
1. draw the structure of each of the common aromatic compounds in Figure 16 (Common benzene derived compounds with various substituents), given their IUPAC-accepted trivial names.
2. write the IUPAC-accepted trivial name for each of the compounds in Figure 16, given the appropriate Kekulé, condensed or shorthand structure.
3. identify the ortho, meta and para positions in a monosubstituted benzene ring.
4. use the ortho/meta/para system to name simple disubstituted aromatic compounds.
5. draw the structure of a simple disubstituted aromatic compound, given its name according to the ortho/meta/para system.
6. provide the IUPAC name of a given aromatic compound containing any number of the following substituents: alkyl, alkenyl or alkynyl groups; halogens; nitro groups; carboxyl groups; amino groups; hydroxyl groups.
7. draw the structure of an aromatic compound containing any number of the substituents listed in Objective 6, above, given the IUPAC name.
8. provide the IUPAC name of a given aromatic compound in which the phenyl group is regarded as a substituent.
9. draw the Kekulé, condensed or shorthand structure of an aromatic compound in which the phenyl group is regarded as a substituent, given its IUPAC name.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• arene
• benzyl group
• phenyl group
Study Notes
You should already know the names and structures of several of the hydrocarbons shown in Figure 15.1. A compound containing a benzene ring which has one or more alkyl substituents is called an arene.
A phenyl group consists of a benzene ring with one of its hydrogens removed.
You should memorize the structures and formulas shown in Figure 16. You will meet these compounds frequently throughout the remainder of this course.
Note that the ortho/meta/para system cannot be used when more than two substituents are present in the benzene ring. The “numbering system” can be used instead of the ortho/meta/para system in most cases when only two substituents are present.
Sources of Aromatic Compounds
Initially aromatic compounds were isolated from coal tar. Coal tar, which is a distillate obtained when heating coal at 1000 oC in the absence of air, is a source of an amazing number of aromatic compounds. Many simple aromatic compounds, some of which includes nitrogen, oxygen, and sulfur, as well as hydrocarbons are obtained.
Some of the Aromatic Compounds Obtained from Coal Tar
Prior to World War II, coal tar was the only important source of aromatic hydrocarbons, but during the war the demand for benzene and toluene, a precursor to the explosive TNT, rose so sharply that other sources had to be found. Today, most of the benzene and almost all of the toluene produced in the United States are derived from petroleum. Although petroleum does contain some aromatic compound, it primarily made up of alkanes of various chain lengths. Aromatic compounds are synthesized from petroleum the by a process referred to in the petroleum industry as catalytic re-forming or hydroforming. This involves heating a C6-C10 alkane fraction of petroleum with hydrogen in the presence of a catalyst to modify the molecular structure of its components. Some amazing transformations take place, and the C6-C7 alkanes can be converted to cycloalkanes, which, in turn, are converted to arenes. Benzene, and methylbenzene (toluene) are produced primarily in this way.
Nomenclature of Mono-Substituted Benzenes
Unlike aliphatic organics, nomenclature of benzene-derived compounds can be confusing because a single aromatic compound can have multiple possible names (such as common and systematic names) be associated with its structure. Common names are often used in the nomenclature of aromatic compounds. IUPAC still allows for some of the more widely used common name to be used. A partial list of these common name is shown in Figure \(2\) and there are numerous others. These common names take the place of the benzene base name. Methylbenzene is commonly known with the base name toluene, hydroxyphenol is known as phenol etc. It is very important to be able to identify these structures as they will be utilized in the nomenclature of more complex compounds.
Mono-substituted benzene rings, with a substituent not on the list above, are named with benzene being the parent name. These compounds are named as such: Name of the substituent + Benzene.
The use of Phenyl and Benzyl in Nomenclature
If the alkyl group attached to the benzene contains seven or more carbons the compounds is named as a phenyl substituted alkane. The name phenyl (C6H5-)is often abbreviated (Ph) and comes from the Greek word pheno which means "I bear light". This name commemorates the fact that benzene was first isolated by Michael Faraday in 1825 from the residue left in London street lamps which burned coal gas. If the alkyl substituent is smaller than the benzene ring (six or fewer carbons), the compound is named as an alkyl-substituted benzene following the rules listed above.
The benzyl group (abbv. Bn), similar to the phenyl group and can be written as C6H5CH2-R, PhCH2-R, or Bn-R. Nomenclature of benzyl group based compounds are very similar to the phenyl group compounds. For example, a chlorine attached to a benzyl group would simply be called benzyl chloride, whereas an OH group attached to a benzyl group would simply be called benzyl alcohol.
Nomenclature of Disubstituted Benzenes
With disubstituted benzenes there are three distinct positional isomers which can occur and must be identified in the compounds name. Although numbering can be used to indicate the position of the two subsituents it is much more common for the compounds to be named using prefixes. These prefixes are italicized and are often abbreviated with a single letter. They are defined as the following:
• ortho- (o-): 1,2- (next to each other in a benzene ring)
• meta- (m): 1,3- (separated by one carbon in a benzene ring)
• para- (p): 1,4- (across from each other in a benzene ring)
Nomenclature of Benzenes with Three or more Substituents
When three or more substituents are present the ortho, meta, para positional prefixes become inadequate and a numbering system for the ring must be applied. Here again it is important to check if any of the substituents are listed in Figure \(2\). If a substituent from Figure \(2\) is present it is given the parent name in the nomenclature. Also, this substituent is given position one in the numbering system. The other substituents are numbered such that they get the lowest possible sum. In the compound's name the subsituents are given their position number and listed alphabetically. Remember that di-, tri, tetra- prefixes are still used to indicate multiple of the same substituent being present but are ignored for alphabetical listing.
Exercises
Exercise \(1\)
(True/False) The compound above contains a benzene ring and thus is aromatic.
Answer
False, this compound does not contain a benzene ring in its structure.
Exercise \(2\)
Benzene unusual stability is caused by how many conjugated pi bonds in its cyclic ring? ____
Answer
3
Exercise \(3\)
Menthol, a topical analgesic used in many ointments for the relief of pain, releases a peppermint aroma upon exposure to the air. Based on this conclusion, can you imply that a benzene ring is present in its chemical structure? Why or why not?
Answer
No, a substance that is fragrant does not imply a benzene ring is in its structure. See camphor example (figure 1)
Exercise \(4\)
Answer
No reaction, benzene requires a special catalyst to be hydrogenated due to its unusual stability given by its three conjugated pi bonds.
Exercise \(5\)
At normal conditions, benzene has ___ resonance structures.
Answer
2
Exercise \(6\)
Which of the following name(s) is/are correct for the following compound?
1. nitrohydride benzene
2. phenylamine
3. phenylamide
4. aniline
5. nitrogenhydrogen benzene
6. All of the above is correct
Answer
b, d
Exercise \(7\)
Convert 1,4-dimethylbenzene into its common name.
Answer
p-Xylene
Exercise \(8\)
TNT's common name is: ______________________________
Answer
2,4,6-trinitrotoluene
Exercise \(9\)
Name the following compound using OMP nomenclature:
Answer
p-chloronitrobenzene
Exercise \(10\)
Draw the structure of 2,4-dinitrotoluene.
Answer
Exercise \(11\)
Name the following compound:
Answer
4-phenylheptane
Example \(12\)
Which of the following is the correct name for the following compound?
1. 3,4-difluorobenzyl bromide
2. 1,2-difluorobenzyl bromide
3. 4,5-difluorobenzyl bromide
4. 1,2-difluoroethyl bromide
5. 5,6-difluoroethyl bromide
6. 4,5-difluoroethyl bromide
a
Exercise \(13\)
(True/False) Benzyl chloride can be abbreviated Bz-Cl.
Answer
False, the correct abbreviation for the benzyl group is Bn, not Bz. The correct abbreviation for Benzyl chloride is Bn-Cl.
Exercise \(14\)
Benzoic Acid has what R group attached to its phenyl functional group?
Answer
COOH
Exercise \(15\)
(True/False) A single aromatic compound can have multiple names indicating its structure.
Answer
True. TNT, for example, has the common name 2,4,6-trinitrotoluene and its systematic name is 2-methyl-1,3,5-trinitrobenzene.
Exercise \(16\)
List the corresponding positions for the OMP system (o-, m-, p-).
Answer
Ortho - 1,2 ; Meta - 1,3 ; Para - 1,4
Exercise \(17\)
A scientist has conducted an experiment on an unknown compound. He was able to determine that the unknown compound contains a cyclic ring in its structure as well as an alcohol (-OH) group attached to the ring. What is the unknown compound?
1. Cyclohexanol
2. Cyclicheptanol
3. Phenol
4. Methanol
5. Bleach
6. Cannot determine from the above information
Answer
The correct answer is f). We cannot determine what structure this is since the question does not tell us what kind of cyclic ring the -OH group is attached on. Just as cyclohexane can be cyclic, benzene and cycloheptane can also be cyclic.
Exercise \(18\)
Which of the following statements is false for the compound, phenol?
1. Phenol is a benzene derived compound.
2. Phenol can be made by attaching an -OH group to a phenyl group.
3. Phenol is highly toxic to the body even in small doses.
4. Phenol can be used as a catalyst in the hydrogenation of benzene into cyclohexane.
5. Phenol is used as an antiseptic in minute doses.
6. Phenol is amongst one of the three common names retained in the IUPAC nomenclature.
Answer
d
Exercise \(19\)
State wither the following is para, meta, or ortho substituted.
Answer
A – meta; B – para; C – ortho
Exercise \(20\)
Name the following compounds.
Answer
a. 1,3-Dibromobenzene
b. 1-phenyl-4-methylhexane
c. 1,4-Dichloro-2,5-dimethylbenzene
d. 2-methyl-1,3,5-trinitrobenzene. (Also known as trinitrotoluene, or TNT)
Exercise \(21\)
Draw the following structures
1. p-chloroiodobenzene
2. m-bromotoluene
3. p-chloroaniline
4. 1,3,5-trimethylbenzene
Answer | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/15%3A_Benzene_and_Aromaticity%3A_Electrophilic_Aromatic_Substitution/15.01%3A_Naming__the__Benzenes.txt |
Objectives
After completing this section, you should be able to
1. compare the reactivity of a typical alkene with that of benzene.
2. Use the heat of hydrogenation data to show that benzene is more stable than might be expected for “cyclohexatriene.”
3. state the length of the carbon-carbon bonds in benzene, and compare this length with those of bonds found in other hydrocarbons.
4. describe the geometry of the benzene molecule.
5. describe the structure of benzene in terms of resonance.
6. describe the structure of benzene in terms of molecular orbital theory.
7. draw a molecular orbital diagram for benzene.
Key Terms
Make certain that you can define, and use in context, the key term below.
• degenerate
Study Notes
You may wish to review Sections 1.5 and 14.1 before you begin to study this section.
Note that the figure showing the molecular orbitals of benzene has two bonding (π2 and π3) and two anti-bonding (π* and π5*) orbital pairs at the same energy levels. Orbitals with the same energy are described as degenerate orbitals.
Structure of Benzene
When benzene (C6H6) was first discovered its low hydrogen to carbon ratio (1:1) led chemists to believe it contained double or triple bonds. Since double and triple bonds rapidly add bromine (Br2), this reaction was applied to benzene. Surprisingly, benzene was entirely unreactive toward bromine. In addition, if benzene is forced to react with bromine through the addition of a catalyst, it undergoes substitution reactions rather than the addition reactions that are typical of alkenes. These experiments suggested that the six-carbon benzene core is unusually stable to chemical modification.
The conceptual contradiction presented by a high degree of unsaturation (low H:C ratio) and high chemical stability for benzene and related compounds remained an unsolved puzzle for many years. Eventually, the presently accepted structure of benzene as a hexagonal, planar ring of carbons with alternating single and double bonds was adopted, and the exceptional chemical stability of this system was attributed to special resonance stabilization of the conjugated cyclic triene. No single structure provides an accurate representation of benzene as it is a combination of two structurally and energetically equivalent resonance forms representing the continuous cyclic conjugation of the double bonds. In the past, the benzene resonance hybrid was represented by a hexagon with a circle in the center to represent the benzene's pi-electron delocalization. This method has largely been abandoned because it does not show the pi electrons contained in benzene. Currently, the structure of benzene is usually represented by drawing one resonance form with the understanding that it does not completely represent the true structure of benzene.
The six-membered ring in benzene is a perfect hexagon with all carbon-carbon bonds having an identical length of 139 pm1. The 139 pm bond length is roughly in between those of a C=C double bond (134 pm) and a C-C single (154 pm) which agrees with the benzene ring being a resonance hybrid made up of 1.5 C-C bonds. Each carbon in the benzene ring is sp2 hybridized which makes all the C-C-C and H-C-C bond angles in benzene 120o and makes the overall molecule planar.
The High Stability of Benzene
In previous polyalkene, examples the electron delocalization described by resonance enhanced the stability of the molecule. However, benzene's stability goes beyond this. Evidence for the enhanced thermodynamic stability of benzene was obtained from measurements of the heat released when double bonds in a six-carbon ring are hydrogenated (hydrogen is added catalytically) to give cyclohexane as a common product. In the following diagram cyclohexane represents a low-energy reference point. Addition of hydrogen to cyclohexene produces cyclohexane and releases heat amounting to 28.6 kcal per mole. If we take this value to represent the energy cost of introducing one double bond into a six-carbon ring, we would expect a cyclohexadiene to release 57.2 kcal per mole on complete hydrogenation, and 1,3,5-cyclohexatriene to release 85.8 kcal per mole.
These heats of hydrogenation would reflect the relative thermodynamic stability of the compounds. In practice, 1,3-cyclohexadiene is slightly more stable than expected, by about 2 kcal, presumably due to conjugation of the double bonds. Benzene, however, is an extraordinary 36 kcal/mole more stable than expected. This sort of stability enhancement is called aromaticity and molecules with aromaticity are called aromatic compounds. Benzene is the most common aromatic compound but there are many others. Aromatic stabilization explains benzene's lack of reactivity compared to typical alkenes.
Atomic Orbitals of Benzene
Also, each of benzene's six carbon atoms are sp2 hybridized and have an unhybrized p orbital perpendicular to plane of the ring. Because each of the six carbon atoms and their corresponding p orbitals are equivalent, it is impossible for them to only overlap with one adjacent p orbital to create three defined double bonds. Instead each p orbital overlaps equally with both adjacent orbitals creating a cyclic overlap involving all six p orbitals. This allows the p orbitals to be delocalized in molecular orbitals that extend all the way around the ring allowing for more overlap than would be obtained from the linear 1,3,5-hexatriene equivalent.
For this to happen, of course, the ring must be planar – otherwise the p orbitals couldn’t overlap properly and benzene is known to be a flat molecule. An electrostatic potential map of benzene, shown below, shows that the pi electrons are evenly distributed around the ring and that every carbon equivalent.
The Molecular Orbitals of Benzene
A molecular orbital description of benzene provides a more satisfying and more general treatment of "aromaticity". We know that benzene has a planar hexagonal structure in which all the carbon atoms are sp2 hybridized, and all the carbon-carbon bonds are equal in length. As shown below, the remaining cyclic array of six p-orbitals ( one on each carbon) overlap to generate six molecular orbitals, three bonding and three antibonding. The plus and minus signs shown in the diagram do not represent electrostatic charge, but refer to phase signs in the equations that describe these orbitals (in the diagram the phases are also color coded). When the phases correspond, the orbitals overlap to generate a common region of like phase, with those orbitals having the greatest overlap (e.g. π1) being lowest in energy. The remaining carbon valence electrons then occupy these molecular orbitals in pairs, resulting in a fully occupied (6 electrons) set of bonding molecular orbitals. It is this completely filled set of bonding orbitals, or closed shell, that gives the benzene ring its thermodynamic and chemical stability, just as a filled valence shell octet confers stability on the inert gases.
To better see source of the stabilizing aromaticity effect created by the cyclic p orbitals of benzene, the molecular orbitals of 1,3,5-hexatriene must be investigated. The molecule 1,3,5-hexatriene contains six p orbitals which all overlap but in a linear fashion. As with benzene, this overlap creates 3 stabilized bonding molecular which are completely filled with six p electron. As expected, the conjugation creates a marked increase of stability in 1,3,5-hexatriene but not as much as in benzene.
The main difference in stability can be seen when comparing the lowest energy molecular orbital of 1,3,5-hexatriene and benzene: pi1. In pi1 molecular orbital of 1,3,5-hexatriene there are 5 stabilizing bonding interactions where there are 6 stabilizing bonding interactions in the pi1 of benzene. The sixth bonding interaction is made possible by benzene's p orbitals being in a ring. Because benzene's pi1 molecular orbital has more stabilizing bonding interactions it is lower in energy than the pi1 molecular orbital of 1,3,5-hexatriene. This gives benzene the additional aromatic stability not seen in the acyclic 1,3,5-hexatriene.
Exercises
Exercise \(1\)
The molecule, pyridine, is planar with bond angles of 120o. Pyridine has many other characteristics similar to benzene. Draw a diagram showing the p orbitals in pyridine and use it to explain its similarity to benzene.
Answer
The nitrogen and each carbon in the pyridine ring is sp2 hybridized. In the bonding picture for pyridine, the nitrogen is sp2-hybridized, with two of the three sp2 orbitals forming sigma overlaps with the sp2 orbitals of neighboring carbon atoms, and the third nitrogen sp2 orbital containing the lone pair electrons. The unhybridized p orbital contains a single electron, which is part of the 6 pi-electron system delocalized around the ring. Pyridine is most likely aromatic which gives it its planar shape and 120o bond angles.
Exercise \(2\)
The molecule shown, p-methylpyridine, has similar properties to benzene (flat, 120° bond angles). Draw the pi-orbitals for this compound.
Answer | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/15%3A_Benzene_and_Aromaticity%3A_Electrophilic_Aromatic_Substitution/15.02%3A%09Structure_and__Resonance_Energy__of__Benzene%3A_A_First__Look_at_Aromaticity.txt |
Objectives
After completing this section, you should be able to
1. compare the reactivity of a typical alkene with that of benzene.
2. Use the heat of hydrogenation data to show that benzene is more stable than might be expected for “cyclohexatriene.”
3. state the length of the carbon-carbon bonds in benzene, and compare this length with those of bonds found in other hydrocarbons.
4. describe the geometry of the benzene molecule.
5. describe the structure of benzene in terms of resonance.
6. describe the structure of benzene in terms of molecular orbital theory.
7. draw a molecular orbital diagram for benzene.
Key Terms
Make certain that you can define, and use in context, the key term below.
• degenerate
Study Notes
You may wish to review Sections 1.5 and 14.1 before you begin to study this section.
Note that the figure showing the molecular orbitals of benzene has two bonding (π2 and π3) and two anti-bonding (π* and π5*) orbital pairs at the same energy levels. Orbitals with the same energy are described as degenerate orbitals.
Structure of Benzene
When benzene (C6H6) was first discovered its low hydrogen to carbon ratio (1:1) led chemists to believe it contained double or triple bonds. Since double and triple bonds rapidly add bromine (Br2), this reaction was applied to benzene. Surprisingly, benzene was entirely unreactive toward bromine. In addition, if benzene is forced to react with bromine through the addition of a catalyst, it undergoes substitution reactions rather than the addition reactions that are typical of alkenes. These experiments suggested that the six-carbon benzene core is unusually stable to chemical modification.
The conceptual contradiction presented by a high degree of unsaturation (low H:C ratio) and high chemical stability for benzene and related compounds remained an unsolved puzzle for many years. Eventually, the presently accepted structure of benzene as a hexagonal, planar ring of carbons with alternating single and double bonds was adopted, and the exceptional chemical stability of this system was attributed to special resonance stabilization of the conjugated cyclic triene. No single structure provides an accurate representation of benzene as it is a combination of two structurally and energetically equivalent resonance forms representing the continuous cyclic conjugation of the double bonds. In the past, the benzene resonance hybrid was represented by a hexagon with a circle in the center to represent the benzene's pi-electron delocalization. This method has largely been abandoned because it does not show the pi electrons contained in benzene. Currently, the structure of benzene is usually represented by drawing one resonance form with the understanding that it does not completely represent the true structure of benzene.
The six-membered ring in benzene is a perfect hexagon with all carbon-carbon bonds having an identical length of 139 pm1. The 139 pm bond length is roughly in between those of a C=C double bond (134 pm) and a C-C single (154 pm) which agrees with the benzene ring being a resonance hybrid made up of 1.5 C-C bonds. Each carbon in the benzene ring is sp2 hybridized which makes all the C-C-C and H-C-C bond angles in benzene 120o and makes the overall molecule planar.
The High Stability of Benzene
In previous polyalkene, examples the electron delocalization described by resonance enhanced the stability of the molecule. However, benzene's stability goes beyond this. Evidence for the enhanced thermodynamic stability of benzene was obtained from measurements of the heat released when double bonds in a six-carbon ring are hydrogenated (hydrogen is added catalytically) to give cyclohexane as a common product. In the following diagram cyclohexane represents a low-energy reference point. Addition of hydrogen to cyclohexene produces cyclohexane and releases heat amounting to 28.6 kcal per mole. If we take this value to represent the energy cost of introducing one double bond into a six-carbon ring, we would expect a cyclohexadiene to release 57.2 kcal per mole on complete hydrogenation, and 1,3,5-cyclohexatriene to release 85.8 kcal per mole.
These heats of hydrogenation would reflect the relative thermodynamic stability of the compounds. In practice, 1,3-cyclohexadiene is slightly more stable than expected, by about 2 kcal, presumably due to conjugation of the double bonds. Benzene, however, is an extraordinary 36 kcal/mole more stable than expected. This sort of stability enhancement is called aromaticity and molecules with aromaticity are called aromatic compounds. Benzene is the most common aromatic compound but there are many others. Aromatic stabilization explains benzene's lack of reactivity compared to typical alkenes.
Atomic Orbitals of Benzene
Also, each of benzene's six carbon atoms are sp2 hybridized and have an unhybrized p orbital perpendicular to plane of the ring. Because each of the six carbon atoms and their corresponding p orbitals are equivalent, it is impossible for them to only overlap with one adjacent p orbital to create three defined double bonds. Instead each p orbital overlaps equally with both adjacent orbitals creating a cyclic overlap involving all six p orbitals. This allows the p orbitals to be delocalized in molecular orbitals that extend all the way around the ring allowing for more overlap than would be obtained from the linear 1,3,5-hexatriene equivalent.
For this to happen, of course, the ring must be planar – otherwise the p orbitals couldn’t overlap properly and benzene is known to be a flat molecule. An electrostatic potential map of benzene, shown below, shows that the pi electrons are evenly distributed around the ring and that every carbon equivalent.
The Molecular Orbitals of Benzene
A molecular orbital description of benzene provides a more satisfying and more general treatment of "aromaticity". We know that benzene has a planar hexagonal structure in which all the carbon atoms are sp2 hybridized, and all the carbon-carbon bonds are equal in length. As shown below, the remaining cyclic array of six p-orbitals ( one on each carbon) overlap to generate six molecular orbitals, three bonding and three antibonding. The plus and minus signs shown in the diagram do not represent electrostatic charge, but refer to phase signs in the equations that describe these orbitals (in the diagram the phases are also color coded). When the phases correspond, the orbitals overlap to generate a common region of like phase, with those orbitals having the greatest overlap (e.g. π1) being lowest in energy. The remaining carbon valence electrons then occupy these molecular orbitals in pairs, resulting in a fully occupied (6 electrons) set of bonding molecular orbitals. It is this completely filled set of bonding orbitals, or closed shell, that gives the benzene ring its thermodynamic and chemical stability, just as a filled valence shell octet confers stability on the inert gases.
To better see source of the stabilizing aromaticity effect created by the cyclic p orbitals of benzene, the molecular orbitals of 1,3,5-hexatriene must be investigated. The molecule 1,3,5-hexatriene contains six p orbitals which all overlap but in a linear fashion. As with benzene, this overlap creates 3 stabilized bonding molecular which are completely filled with six p electron. As expected, the conjugation creates a marked increase of stability in 1,3,5-hexatriene but not as much as in benzene.
The main difference in stability can be seen when comparing the lowest energy molecular orbital of 1,3,5-hexatriene and benzene: pi1. In pi1 molecular orbital of 1,3,5-hexatriene there are 5 stabilizing bonding interactions where there are 6 stabilizing bonding interactions in the pi1 of benzene. The sixth bonding interaction is made possible by benzene's p orbitals being in a ring. Because benzene's pi1 molecular orbital has more stabilizing bonding interactions it is lower in energy than the pi1 molecular orbital of 1,3,5-hexatriene. This gives benzene the additional aromatic stability not seen in the acyclic 1,3,5-hexatriene.
Exercises
Exercise \(1\)
The molecule, pyridine, is planar with bond angles of 120o. Pyridine has many other characteristics similar to benzene. Draw a diagram showing the p orbitals in pyridine and use it to explain its similarity to benzene.
Answer
The nitrogen and each carbon in the pyridine ring is sp2 hybridized. In the bonding picture for pyridine, the nitrogen is sp2-hybridized, with two of the three sp2 orbitals forming sigma overlaps with the sp2 orbitals of neighboring carbon atoms, and the third nitrogen sp2 orbital containing the lone pair electrons. The unhybridized p orbital contains a single electron, which is part of the 6 pi-electron system delocalized around the ring. Pyridine is most likely aromatic which gives it its planar shape and 120o bond angles.
Exercise \(2\)
The molecule shown, p-methylpyridine, has similar properties to benzene (flat, 120° bond angles). Draw the pi-orbitals for this compound.
Answer | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/15%3A_Benzene_and_Aromaticity%3A_Electrophilic_Aromatic_Substitution/15.03%3A_Pi_Molecular__Orbitals__of_Benzene.txt |
Objectives
After completing this section, you should be able to
1. determine whether an unknown compound contains an aromatic ring by inspection of its infrared spectrum, given a table of characteristic infrared absorptions.
2. state the approximate chemical shift of aryl protons in a proton NMR spectrum.
3. explain why signals resulting from the presence of aryl protons are found downfield from those caused by vinylic protons in a proton NMR spectrum.
4. propose possible structures for an unknown aromatic compound, given its proton NMR spectrum, other spectroscopic data (such as a 13C NMR or infrared spectrum), or both.
Key Terms
Make certain that you can define, and use in context, the key term below.
• ring current
Study Notes
It is not necessary that you memorize detailed spectroscopic data. In the laboratory, on assignments and when writing examinations, you will be provided with a table of characteristic infrared absorptions to assist you in interpreting infrared spectra.
The important points to note about the proton NMR of aromatic compounds are the approximate chemical shifts of such protons and the complex splitting pattern that is sometimes observed. You are advised not to spend too long trying to understand why the signal for an aryl proton is found downfield from the signal for a vinylic proton. In general, we want you to be able to interpret NMR spectra, and leave the underlying theory for subsequent chemistry courses.
The chemical shifts of aromatic protons
Some protons resonate much further downfield than can be accounted for simply by the deshielding effect of nearby electronegative atoms. Vinylic protons (those directly bonded to an alkene carbon) and aromatic (benzylic) protons are dramatic examples.
We'll consider the aromatic proton first. Recall that in benzene and many other aromatic structures, a sextet of p electrons is delocalized around the ring. When the molecule is exposed to B0, these p electrons begin to circulate in a ring current, generating their own induced magnetic field that opposes B0. In this case, however, the induced field of the p electrons does not shield the benzylic protons from B0 as you might expect– rather, it causes the protons to experience a stronger magnetic field in the direction of B0 – in other words, it adds to B0 rather than subtracting from it.
To understand how this happens, we need to understand the concept of diamagnetic anisotropy (anisotropy means `non-uniformity`). So far, we have been picturing magnetic fields as being oriented in a uniform direction. This is only true over a small area. If we step back and take a wider view, however, we see that the lines of force in a magnetic field are actually anisotropic. They start in the 'north' direction, then loop around like a snake biting its own tail.
If we are outside the ring in the figure above, we feel a magnetic field pointing in a northerly direction. If we are inside the ring, however, we feel a field pointing to the south.
In the induced field generated by the aromatic ring current, the benzylic protons are outside the ring – this means that the induced current in this region of space is oriented in the same direction as B0.
In total, the benzylic protons are subjected to three magnetic fields: the applied field (B0) and the induced field from the electrons pointing in one direction, and the induced field of the non-aromatic electrons pointing in the opposite (shielding) direction. The end result is that benzylic protons, due to the anisotropy of the induced field generated by the ring current, appear to be highly de-shielded. Their chemical shift is far downfield, in the 6.5-8 ppm region.
The presence of anisotropy effects is often used to indicate aromaticity in a molecule. An example is the nonaromatic 8 pi electron cyclooctatetraene. In order to avoid anti-aromatic destabilization, the molecule takes a non-planar conformation which prevents conjugation and ring currents. Consequently, the molecule's protons absorbed at 5.8 ppm which is within the alkene region of 1H NMR.
Exercise \(1\)
The 1H-NMR spectrum of [18] annulene has two peaks, at 8.9 ppm and -1.8 ppm (upfield of TMS!) with an integration ratio of 2:1. Explain the unusual chemical shift of the latter peak.
Answer
The molecule contains two groups of equivalent protons: the twelve pointing to the outside of the ring, and the six pointing into the center of the ring. The molecule is aromatic, as evidenced by the chemical shift of the ‘outside’ protons’ (and the fact that there are 18 pelectrons, a Hückel number). The inside protons are shielded by the induced field of the aromatic ring current, because inside the ring this field points in the opposite direction of B0. This is the source of the unusually low (negative relative to TMS) chemical shift for these protons.
Characteristic 1H NMR Absorptions of Aromatic Compounds
Protons directly attached to an aromatic ring, commonly called aryl protons, show up about 6.5-8.0 PPM. This range is typically called the aromatic region of an 1H NMR spectrum. Protons on carbons directly bonded to an aromatic ring, called benzylic protons, show up about 2.0-3.0 PPM.
'
For the 1H NMR spectrum of p-bromotoluene, the absorption for the benzylic protons appears as a strong singlet at 2.28 ppm. Due to the molecule's symmetry, the aryl protons appear as two doublets at 6.96 & 7.29 ppm.
Characteristic 13C NMR Absorptions of Aromatic Compounds
Carbons in an aromatic ring absorb in the range of 120-150 ppm in a 13C NMR spectrum. This is virtually the same range as nonaromatic alkenes (110-150 ppm) so peaks in this region are not definitive proof of a molecule's aromaticity.
Due to the decoupling in 13C NMR, the number of absorptions due to aromatic carbons can easily be observed. This can be used to determine the relative positions (ortho, meta, or para) for di-substituted benzenes. Peak assignments can be simplified by noting that 13C peaks tend to be larger if two carbons contribute to the absorption.
Symmetrical Di-substituted Benzenes
Benzenes substituted with two identical groups have a relatively high amount of symmetry. In all three configurations, there is a plane of symmetry which reduces the number of distinct aryl carbon absorptions to less than six. The ortho configuration has a plane of symmetry which mirrors each carbon in the benzene ring causing only three 13C aryl absorptions to occur. The meta configuration's plane of symmetry mirrors two carbons of the benzene ring allowing for four aryl absorptions to occur. The para configuration actually has two planes of symmetry (one vertical and one horizontal on the structure below) through the benzene ring, which only allows two distinct aryl absorptions to occur.
Asymmetrical Di-substituted Benzenes
The lack of a plane of symmetry in asymmetrical di-substituted benzenes makes each carbon in the ortho and meta configuration unique. Consequently, their 13C NMR spectra show six arene absorptions. However, the para configuration has a plane of symmetry drawn through the two substituents which mirrors two carbons of the benzene ring causing only 4 arene absorptions to appear.
Charateristic IR Absorption of Benzene Derivatives
Arenes have absorption bands in the 650-900 cm−1 region due to bending of the C–H bond out of the plane of the ring. The exact placement of these absorptions can indicate the pattern of substitution on a benzene ring. However, this is beyond the scope of introductory organic chemistry. Arenes also possess a characteristic absorption at about 3030-3100 cm−1 as a result of the aromatic C–H stretch. It is somewhat higher than the alkyl C–H stretch (2850–2960 cm−1), but falls in the same region as olefinic compounds. Two bands (1500 and 1660 cm−1) caused by C=C in plane vibrations are the most useful for characterization as they are intense and are likely observed.
In aromatic compounds, each band in the spectrum can be assigned:
• C–H stretch from 3100-3000 cm-1
• overtones, weak, from 2000-1665 cm-1
• C–C stretch (in-ring) from 1600-1585 cm-1
• C–C stretch (in-ring) from 1500-1400 cm-1
• C–H "oop" from 900-675 cm-1
Note that this is at slightly higher frequency than is the –C–H stretch in alkanes. This is a very useful tool for interpreting IR spectra. Only alkenes and aromatics show a C–H stretch slightly higher than 3000 cm-1.
Ultraviolet Spectroscopy
The presence of conjugated π bonds makes aromatic rings detectable in a UV/Vis spectrum. Benzene primarily absorbs through a be a π-π* transition over the range from 160-208 nm with a λ max value of about 178 nm.
Benzene shows a less intense absorption in 230-276 nm range. These absorptions are due to vibrational fine structure often displayed by rigid molecules such as benzene and other aromatic compounds. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/15%3A_Benzene_and_Aromaticity%3A_Electrophilic_Aromatic_Substitution/15.04%3A_Spectral__Characteristics_of_the__Benzene__Ring.txt |
Objectives
After completing this section, you should be able to draw the resonance contributors for polycyclic aromatic compounds, such as naphthalene, anthracene, etc.
Key Terms
Make certain that you can define, and use in context, the key term below.
• polycyclic aromatic compounds
Study Notes
As their name indicates, polycyclic aromatic hydrocarbons are aromatic hydrocarbons which contain more than one benzenoid (i.e., benzene-like) ring. This section deals only with those compounds in which the benzenoid rings are fused together; in other words, compounds in which at least one carbon-carbon bond is common to two aromatic rings. Another type of polycyclic aromatic hydrocarbon contains two or more benzenoid rings joined by a carbon-carbon single bond. The simplest compound of this type is biphenyl, the compound from which PCBs (polychlorinated biphenyls) are derived.
Polycyclic Aromatic Compounds
Hückel's 4n +2 rule for aromaticity does not only apply to mono-cyclic compounds. Benzene rings may be joined together (fused) to give larger polycyclic aromatic compounds. A few examples are drawn below, together with the approved numbering scheme for substituted derivatives. The peripheral carbon atoms (numbered in all but the last three examples) are all bonded to hydrogen atoms. Unlike benzene, all the C-C bond lengths in these fused ring aromatics are not the same, and there is some localization of the pi-electrons. The six benzene rings in coronene are fused in a planar ring; whereas the six rings in hexahelicene are not joined in a larger ring, but assume a helical turn, due to the crowding together of the terminal ring atoms (in the structure below, note that the top right and center right rings are not attached to one another). This helical configuration renders the hexahelicene molecule chiral, and it has been resolved into stable enantiomers.
Figure 2: Examples of Polycyclic Aromatic Hydrocarbons (PAHs).
Probably the most well know polycyclic aromatic compound which only contains carbon is naphthalene (C10H8). Naphthalene shares many of the same characteristic as benzene. Naphthalene is cyclic and known to be flat. Each carbon in naphthalene is sp2 hybridized so it is completely conjugated. The true structure of naphthalene is a combination of three resonance hybrids.
Naphthalene
Heat of hydrogenation experiments with naphthalene shows an unusual "aromatic" stabilization energy. Also, naphthalene prefers to react with electrophiles to give substitution products rather than the typical double bond addition products.
Every carbon in naphthalene is sp2 hybridized so there is conjugated p orbital overlap over the entire ring system. The electrostatic potential map shows that pi electrons in naphthalene are evenly distributed making each carbon equivalent.
Lastly, naphthalene has 10 pi electrons which fulfills Hückel's rule. The importance of the 4n + 2 rule can be seen when considering the molecular orbital diagram of naphthalene. Naphthalene has 10 p orbitals which is 4 more than benzene. In the molecular orbital diagram of naphthalene, the 4 additional p orbitals become 2 bonding orbitals and two anti-bonding orbitals. The two additional bonding orbitals require 4 additional pi electrons to be filled so naphthalene needs 10 pi electrons in total to fill all of the bonding molecular orbitals.
Polycyclic Aromatic Heterocycles
There is a wide variety of polycyclic aromatic heterocycles, many of which contain nitrogen, oxygen, or sulfur. Some of the biologically important structures, are the nitrogen containing polycyclic aromatic heterocycles indole, quinoline, isoquinoline, and purine which are all polycyclic aromatic heterocycles commonly found in nature. Notice that these compound all have 10 pi electrons.
Indole, quinoline, and isoquinoline all contain a hetrocyclic ring fused to benzene. Purine is made up to two heterocyclic rings, imidazole and pyrimidine, fused together. Quinoline is found in the antimalarial drug quinine. Indole is found in the neurotransmitter serotonin. The purine ring structure is found in adenine and guanine, two important parts of DNA and in the stimulant caffeine.
Exercises
Exercise \(1\)
1) The following molecule is an isomer of naphthalene. Is it aromatic? Draw a resonance structure for it.
2) The following molecule is adenine. It has a purine core. Of the nitrogen in the core, how many electrons are donated into the pi system?
Answer
1) It has 10 pi electrons which follows the 4n+2 rule making it aromatic.
2) There is only one nitrogen of the core that contributes a set of lone pair electrons as 2 pi electrons (in red). All of the other nitrogens are sp2 hybrized and contribute 1 pi electron each. In total, the core is aromatic with 10 electrons in the pi-system.
Exercise \(2\)
This is an isomer of naphthalene. Is it aromatic? Draw a resonance structure for it.
Answer
Yes, it is aromatic. 4n+2 pi-electrons.
Exercise \(3\)
The following molecule is adenine. It has a purine core. Of the nitrogen in the core, how many electrons are donated into the pi system?
Answer
There is only one nitrogen of the core that contributes to the pi-system (in red). With this one lone pair the core is aromatic with 10 electrons in the pi-system.
If we extend the structure of corannulene by adding similar cycles of five benzene rings, the curvature of the resulting molecule should increase, and eventually close into a sphere of carbon atoms. The archetypical compound of this kind (C60) has been named buckminsterfullerene because of its resemblance to the geodesic structures created by Buckminster Fuller. It is a member of a family of similar carbon structures that are called fullerenes. These materials represent a third class of carbon allotropes. Alternating views of the C60 fullerene structure are shown on the right, together with a soccer ball-like representation of the 12 five and 20 six-membered rings composing its surface. Precise measurement by Atomic Force Microscopy (AFM) has shown that the C-C bond lengths of the six-membered rings are not all equal, and depend on whether the ring is fused to a five or six-membered beighbor. By clicking on this graphic, a model of C60 will be displayed.
Although C60 is composed of fused benzene rings its chemical reactivity resembles that of the cycloalkenes more than benzene. Indeed, exposure to light and oxygen slowly degrade fullerenes to cage opened products. Most of the reactions thus far reported for C60 involve addition to, rather than substitution of, the core structure. These reactions include hydrogenation, bromination and hydroxylation. Strain introduced by the curvature of the surface may be responsible for the enhanced reactivity of C60.
. Larger fullerenes, such as C70, C76, C82 & C84have ellipsoidal or distorted spherical structures, and fullerene-like assemblies up to C240 have been detected. A fascinating aspect of these structures is that the space within the carbon cage may hold atoms, ions or small molecules. Such species are called endohedral fullerenes. The cavity of C60 is relatively small, but encapsulated helium, lithium and atomic nitrogen compounds have been observed. Larger fullerenes are found to encapsulate lanthanide metal atoms.
Interest in the fullerenes has led to the discovery of a related group of carbon structures referred to as nanotubes. As shown in the following illustration, nanotubes may be viewed as rolled up segments of graphite. The chief structural components are six-membered rings, but changes in tube diameter, branching into side tubes and the capping of tube ends is accomplished by fusion with five and seven-membered rings. Many interesting applications of these unusual structures have been proposed. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/15%3A_Benzene_and_Aromaticity%3A_Electrophilic_Aromatic_Substitution/15.05%3A_Polycyclic__Aromatic__Hydrocarbons.txt |
Objectives
After completing this section, you should be able to
1. define aromaticity in terms of the Hückel 4n + 2 rule.
2. use the Hückel 4n + 2 rule to determine whether or not a given polyunsaturated cyclic hydrocarbon should exhibit aromatic properties.
3. describe the difference in properties between an aromatic hydrocarbon, such as benzene, and a non-aromatic polyunsaturated cyclic hydrocarbon, such as cyclobutadiene or cyclooctatetraene.
4. draw molecular orbital diagrams for aromatic species, such as benzene, the cyclopentadienyl anion and pyridine, and compare these diagrams with those obtained for non-aromatic species, such as cyclobutadiene and the cyclopentadienyl cation.
Study Notes
The following mnemonic device will help you establish the approximate energy levels for the molecular orbitals of various organic ring systems.
Whatever the size of the ring, place one point of the ring down to the bottom. The corners of the ring, where the carbons are located, will roughly approximate the location and pattern of the molecular orbital energy levels. Cut the ring exactly in half. The energy levels in the top half will be anti-bonding (Ψ*) orbitals and those in the bottom will be bonding (Ψ) orbitals. If the carbons fall directly in the centre of the ring (e.g., four-membered rings) the energy levels there are non-bonding.
cyclopropenyl ring (three-membered ring)
cyclobutadienyl ring (four-membered ring)
cyclopentadienyl ring (five-membered ring)
In 1931, German chemist and physicist Erich Hückel proposed a theory to help determine if a planar ring molecule would have aromatic properties. His rule states that if a cyclic, planar molecule has $4n+2$ $π$ electrons, it is considered aromatic. This rule would come to be known as Hückel's Rule.
Four Criteria for Aromaticity
When deciding if a compound is aromatic, go through the following checklist. If the compound does not meet all the following criteria, it is likely not aromatic.
1. The molecule is cyclic (a ring of atoms)
2. The molecule is planar (all atoms in the molecule lie in the same plane)
3. The molecule is fully conjugated (p orbitals at every atom in the ring)
4. The molecule has 4n+2 π electrons (n=0 or any positive integer)
Counting π Electrons?
Perhaps the toughest part of Hückel's Rule is figuring out which electrons in the compound are actually π electrons. Once this is figured out, the rule is quite straightforward. Pi electrons lie in the p orbitals and sp2 hybridized atoms have 1 of these p orbitals each. When looking at a C=C double bond we know that there is one sigma bond and one pi bond. The pi bond is formed by the overlap of 1 p orbital from each carbon in the double bond. Each p orbital contains one electron so when the two p orbitals overlap the two electrons become a pi bond and are thus called pi electrons. Because aromaticity deals directly with double bonds and conjugation it is simpler to just count the number pi electrons in a compound. Each double bond (π bond) always contributes 2 π electrons. Benzene has 3 double bonds, so it has 6 π electrons.
Why 4n+2 π Electrons?
According to Hückel's Molecular Orbital Theory, a compound is particularly stable if all of its bonding molecular orbitals are filled with paired electrons. This is true of aromatic compounds, meaning they are quite stable. With aromatic compounds, 2 electrons fill the lowest energy molecular orbital, and 4 electrons fill each subsequent energy level (the number of subsequent energy levels is denoted by n), leaving all bonding orbitals filled and no anti-bonding orbitals occupied. This gives a total of 4n+2 $\pi$ electrons. You can see how this works with the molecular orbital diagram for the aromatic compound, benzene, below. Benzene has 6 $\pi$ electrons. Its first 2 $\pi$ electrons fill the lowest energy orbital, and it has 4 $\pi$ electrons remaining. These 4 fill in the orbitals of the succeeding energy level. Notice how all of its bonding orbitals are filled, but none of the anti-bonding orbitals have any electrons.
To apply the 4n+2 rule, first count the number of π electrons in the molecule. Then, set this number equal to 4n+2 and solve for n. If is 0 or any positive integer (1, 2, 3,...), the rule has been met. For example, benzene has six $\pi$ electrons:
\begin{align} 4n + 2 &= 6 \ 4n &= 4 \ n &= 1 \end{align} \nonumber
For benzene, we find that $n=1$, which is a positive integer, so the rule is met.
With Hückel's theory we can assume that if a molecule meets the other criteria for aromaticity and also has 2, 6, 10, 14, 18 ect. pi electrons it will most likely be aromatic.
Antiaromaticity
More than 100 years ago, chemists recognized the possible existence of other conjugated cyclic polyalkenes, which at least superficially would be expected to have properties like benzene. The most interesting of these are cyclobutadiene, whose shape and alignment of p orbitals suggested it should have substantial electron-delocalization energies.
Cyclobutadiene
However, cyclobutadiene was found to be an extremely unstable molecules. In fact, cyclobutadiene is even more reactive than most alkenes. The synthesis of cyclobutadiene eluded chemists for almost 100 years. As more work was done, it became increasingly clear that the molecule, when formed in reactions, was immediately converted to something else. Finally, cyclobutadiene was captured in an essentially rigid matrix of argon at 8K. On warming to even 35K , it dimerizes through a Diels Alder reaction to yield a tricyclicdiene.
Due to the square ring, cyclobutadiene was expected to have some degree of destabilization associated with ring strain. However, estimations of the strain energies, though substantial, did not account for cyclobutadiene's high degree of instability. Also, why was it not being stabilized through cyclic conjugation in the same way as benzene. The answer can be seen in the molecular orbitals.
We considering the molecular orbitals diagram of the analogous 1,3-butadiene, the four 2p atomic orbitals combine to form four pi molecular orbitals of increasing energy. Two bonding pi orbitals and two antibonding pi* orbitals. The 4 pi electrons of 1,3-butadiene completely fill the bonding molecular orbitals giving is the additional stability associated with conjugated double bonds.
However, when placed into a ring the molecular orbitals undergo a significant change. The four p orbitals of cyclobutadiene combine to form the following 4 molecular orbitals:
• a bonding molecular orbital which is lower in energy than the atomic orbitals
• two degenerate non-bonding molecular orbitals which are of equivalent energy to the atomic orbitals
• one antibonding molecular orbital which is higher in energy
The non-bonding orbitals represent that there is no direct interaction between adjacent atoms. When adding cyclobutadiene's 4 pi electrons to the molecular orbital diagram, the bonding orbital is filled and both non-bonding orbitals are singly occupied. If cyclobutadiene's double bonds were delocalized, all the pi electrons would be in low energy bonding orbitals. However, only two of the pi electrons are in bonding orbitals; the other two are non-bonding. In addition, the molecular orbital diagram shows that two of the electrons are unpaired, a situation called a triplet state, which usually makes organic molecules very reactive. For cyclobutadiene cyclic conjugation has made the molecule less stable. This unexpected instability in 4n π-electron cyclic conjugated compounds is termed antiaromaticity and the compounds are called antiaromatic.
Antiaromaticity gives cyclobutadiene some interesting structural features. Cyclobutadiene's single and double bonds have different bond lengths, 158 pm and 135 pm respectively which give it a rectangular shape. If the double bonds were conjugated, there would only be one average bond length, like benzene, and shape would be square. In fact, cyclobutadiene's double bonds are not conjugated but locked into position.
Just like 4n + 2 rule with aromaticity, compounds which are flat, cyclic, have a p orbital at every member of the ring, and 4n pi electrons should be antiaromatic. Another potentially antiaromatic molecule is 1,3,5,7-cyclooctatetraene.
Cyclooctatetraene was first synthesized in 1911 by a German chemist, R. Willstatter (Nobel Prize 1915), who reported an extraordinary thirteen-step synthesis of cyclooctatetraene from a rare alkaloid called pseudopelletierine isolated from the bark of pomegranate trees. However, during the Second World War, the German chemist W. Reppe found that cyclooctatetraene can be made in reasonable yields by the tetramerization of ethyne under the influence of a nickel cyanide catalyst:
If cyclooctatetraene was flat it would have all the requirements to be antiaromatic. It is cyclic, has a p orbital in every member of the ring, and has 8 pi electrons. When looking at the molecular orbital diagram of cyclooctatetraene we see it shares certain characteristics in common with cyclobutadiene. Two of the molecular orbitals are degenerate non-bonding orbitals. When the molecular orbitals are fill with cyclooctatetraene's 8 pi electrons the last two electrons are unpaired in the two nonbonding orbitals creating a triplet state. Based off the molecular orbital diagram cyclooctatetraene should be antiaromatic.
However, cyclooctatetraene is easily prepared and relatively stable. Also, it undergoes addition reactions typical of alkenes. In reality, cyclooctatetraene is not flat but has a tub-like shape. By assuming this shape, the p orbtials between double bond are out of alignment for overlap. Because the double bonds are not conjugated, cyclooctatetraene escapes the destabilizing effects of antiaromaticity. This makes its its pi bonds react like 'normal' alkenes. Because cyclooctatetraene is not flat nor conjugated it is properly defined as non-aromatic. In general, if an antiaromatic compound has the ability to form a non-planar shape it will do so to avoid destabilization by becoming non-aromatic.
Determining if a Compound is Aromatic, Antiaromatic, or Nonaromatic
To make this determination it is important to first as if the compound has the possibility of cyclic conjugation. This requires that the compound be cyclic and have a p orbital at every atom in the ring. If the compound does not have both of these criteria it cannot be aromatic or antiaromatic and must therefore be nonaromatic. Next the number of pi electrons in the ring is determined to see if it follows the count of aromaticity (4n + 2) or antiaromaticity (4n). Before making the final determination, it is vital to know the actual geometry of the molecule. Despite the number of pi electrons, a compound cannot be aromatic or antiaromatic if its geometry is not planar to allow for p orbital overlap.
A Common Misconception
A very common misconception is that hybridization can be used to predict the geometry, or that hybridization somehow involves an energy cost associated with 'promoting' electrons into the hybrid orbitals. This is entirely wrong. Hybridization is always determined by geometry. You can only assign hybridization states to an atom if you already know its geometry, based on some experimental or theoretical evidence. The geometry of the oxygen in furan is trigonal planar and therefore the hybridization must be sp2.
The specific rule is that if you have an sp2 conjugated system, the lone pair will be involved if it makes the system more stable. In this case, conferring Hückel 4n+2 aromaticity. For furan with two lone pairs on the oxygen atom, if we count electrons from the carbon atoms, we have 4 (one per carbon). So adding two electrons from one of the lone pairs will give 6 = 4(1)+2, so Hückel rule is applicable and furan is aromatic.
Exercise $1$
Using the criteria for aromaticity, determine if the following molecules are aromatic:
Answer
a) Aromatic - only 1 of S's lone pairs counts as π electrons, so there are 6 π electrons, n=1
b) Not aromatic - not fully conjugated, top C is sp3 hybridized
c) Not aromatic - top C is sp2 hybridized, but there are 4 π electrons, n=1/2
d) Aromatic - N is using its 1 p orbital for the electrons in the double bond, so its lone pair of electrons are not π electrons, there are 6 π electrons, n=1
e) Aromatic - there are 6 π electrons, n=1
f) Not aromatic - all atoms are sp2 hybridized, but only 1 of S's lone pairs counts as π electrons, so there 8 π electrons, n=1.5
g) Not aromatic - there are 4 π electrons, n=1/2
h) Aromatic - only 1 of N's lone pairs counts as π electrons, so there are 6 π electrons, n=1
i) Not aromatic - not fully conjugated, top C is sp3 hybridized
j) Aromatic - O is using its 1 p orbital for the elections in the double bond, so its lone pair of electrons are not π electrons, there are 6 π electrons, n=1
Exercise $2$
To be aromatic, a molecule must be planar conjugated, and obey the 4n+2 rule. The following is the following molecule aromatic?
Answer
No, it is not. It does not obey the 4n+2 rule. Also it is not planar. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/15%3A_Benzene_and_Aromaticity%3A_Electrophilic_Aromatic_Substitution/15.06%3A_Other__Cyclic__Polyenes%3A__Huckel_%27s_Rule.txt |
Objectives
After completing this section, you should be able to
1. use the Hückel 4n + 2 rule to explain the stability of the cyclopentadienyl anion, the cycloheptatrienyl cation and similar species.
2. use the Hückel 4n + 2 rule to determine whether or not a given unsaturated cyclic hydrocarbon anion or cation is aromatic.
3. draw the resonance contributors for the cyclopentadienyl anion, cation and radical, and similar species.
Aromatic Ions
As previously, noted a ring must be fully conjugated to have the potential to be aromatic. This means that every atom in the ring must have a p orbital which can overlap with adjacent p orbitals. Until now the atoms providing the p orbitals have been neutral sp2 hybridized carbons, however it is also possible for sp2 hybridized carbons to have a charge. There are several examples of cationic and anionic compounds with unexpected stabilities that suggest that they are aromatic. It is important to understand how charged carbons in these compounds will affect the determination of aromaticity.
There are two main situations which need to be considered: The conjugation of a carbocation and the conjugation of a carbanion. A carbocation carbon is surrounded by three electron groups giving it sp2 hybridization. The remaining unhybridized p orbital holds the carbocation's positive charge and is vacant of pi electrons. Although a carbocation is capable of extending conjugation it does not add to the compound's pi electron count.
A carbanion carbon is surrounded by four electron groups and would normally be sp3 hybridized. However, to obtain the stabilizing effects of conjugation, carbanion carbons can becomes sp2 hybridized putting the set of lone pair electrons into the unhybridized p orbital. Because the carbanion's p orbital contains two electrons in the form of a set of lone pair electrons, it increases a compound's pi electron count by 2.
Cyclopentadiene Ion
One of the most well know examples of an aromatic ion is the 1,3-cyclopentadiene ion. 1,3-Cyclopentadiene is nonaromatic due to the presence of an intervening sp3 hybridized -CH2- carbon atom which prevents pi electrons from delocalizing about the entire ring. Also, it only has 4 pi electrons which does not follow Hückel's 4n + 2 rule. However, if a proton is removed form the CH2 group to form the cyclopentadienyl anion, the carbon atom becomes sp2 hybridized and the two electrons of the resulting lone pair occupy the newly produced p orbital. This increases the number of pi electrons in the cyclopentadienyl anion to 6 which follows the 4n +2 rule. Moreover, this new p orbital overlaps with the p orbitals already present allowing for cyclic delocalization of pi electrons about the entire ring.
The lone pair electrons and the negative charge are conjugated about the entire ring making each carbon in the cyclopentadienyl anion equivalent with 1/5 the negative charge. The resonance hybrid can be shown by drawing a series of five resonance form. Also, the electrostatic potential map of the cyclopentadienyl anion show the negative charge, seen in red/yellow is distributed over the entire ring. The fact that the red/yellow color is held in the center of the ring indicates that the extra electrons of the ion are involved in the aromatic p-electron system.
The reason why the 4n +2 rule still works for a 5 p orbital ring system can be seen by looking the molecular orbital diagram of the cyclopentadienyl anion. As discussed in Section 15.3, the molecular orbital diagram of a 5 p orbital system is made up of 3 bonding MO's and 2 antibonding MO's. The 6 pi electrons gained by forming an anion is enough to completely fill the bonding MO's in the diagram giving the cyclopentadienyl anion aromaticity.
One of the effects of the aromaticity of the cyclopentadienyl anion is that the acidity of 1,3-cyclopentadiene is unusually strong. As previously discussed, stabilizing the conjugate base increases the acidity of the corresponding acid. In this case, the conjugate base of 1,3-cyclopentadiene, the cyclopentadienyl anion, is stabilized through aromaticity. This makes 1,3-cyclopentadiene one of the most acidic hydrocarbons known with a pKa of 16. This is almost 1030 times more acidic than cyclopentane. Because of its acidity, cyclopentadiene can be deprotonated by moderately strong bases such as NaOH.
Tropylium ion
The molecule 1,3,5-cycloheptatriene has six pi electrons but is nonaromatic due the presence of an sp3 hybridized -CH2- group which prevents cyclic delocalization. When 1,3,5-cycloheptatriene is reacted with a reagent that can remove a hydride ion (H:-), the 1,3,5-cycloheptatrienyl cation, which is commonly known as the tropylium cation, is formed with unexpected ease. Despite the presence of an electron deficient carbocation, the tropylium cation, is unusually stable and can be isolated as a salt.
Removal of a hydride from the -CH2- group in 1,3,5-cycloheptatriene creates an sp2 hybridized carbocation with a vacant p orbital. The new p orbital allows for cyclic conjugation to occur among the seven p orbitals in the tropylium cation. The vacant p orbital does not change the pi electron count so the tropylium cation has 6 pi electrons which obeys the 4n + 2 rule for aromaticity.
The molecular orbital diagram for the 7 p orbitals in the tropylium cation has three bonding MO's and 4 antibonding MO's. The tropylium cation's 6 pi electrons completely fill the bonding molecular orbitals which is consistent with the tropylium cation being aromatic and therefore unusually stable.
As predicted with aromaticity, the positive charge is completely conjugated about the entire ring giving each carbon +1/7 charge. The true resonance hybrid can be depicted by drawing a series of seven resonance form. Also, the electrostatic potential map of the tropylium cation shows the positive charge is evenly distributed over the entire ring. The equivalency of the carbons in the seven membered ring is experimentally supported by an 1H NMR spectrum of the tropylium cation which contains one peak showing that all seven protons are equivalent.
Antiaromatic Ions
In a similar fashion, cyclically conjugated ions with 4n pi electrons can be predicted to be antiaromatic and therefore highly unstable. An excellent example is the cyclopentadienyl cation. Above, the cyclopentadienyl anion was shown to be aromatic, however, the formation of a carbocation produces a different result. Although the vacant p orbital provided by the carbocation allows for cyclic conjugation to occur the compound only has 4 pi electrons.
After placing these 4 pi electrons into the molecular orbital for a cyclic 5 p orbital species the bonding molecular orbitals remain unfilled. Two of the pi electrons are unpaired in degenerate molecular orbitals creating the highly unstable triplet state. As predicted by Hückel's Rule, the cyclopentadienyl cation has 4n pi electrons and should be anitaromatic and very unstable.
Although there is some discussion as to whether the cyclopentadienyl cation is truly antiaromatic, there is experimental evidence which shows it is usually unstable. In particular, the compound 2,4-cyclopentadien-1-ol is usually resistant to SN1 reactions with acid halides. The carbocation intermediate created during the mechanism of this reaction would be expected to easily form due to the resonance stabilization provided by the two double bonds. Because the carbocation intermediate is antiaromatic it does not form so the reaction does not occur.
Exercise \(1\)
1) Are all bonds equivalent cyclopentadienyl anion? How many lines (signals) would you expect to see in a H1 and C13 NMR spectrum?
2) The following reaction occurs readily. Propose a reason why this occurs?
3) Is the cyclopropenium ion aromatic? Use the 4n + 2 rule and the MO diagram for a 3 p orbital ring system to explain your answer.
Answer
1) Due to the cyclopentadienyl anion being fully conjugated a protons and carbons are equivalent. Both the 1H and 13C NMR spectrum would only have one signal.
2) The addition of two electrons gives the ring system 10 pi electrons which follows the 4n+2 rule making the dianion aromatic. When looking at the MO diagram for an 8 p orbital ring system, 10 pi electrons completely fills the three bonding MO's and the two nonbonding MO's allowing for aromaticity.
3) The cyclopropenium ion has 2 pi electrons which follows the 4n + 2 rule when n = 0. When considering the molecular orbital diagram for a 3 p orbital ring system the is one bonding MO and two antibonding MO's. The two pi electrons completely fill the bonding MO allowing for aromaticity.
Exercises
Exercise \(2\)
Draw the resonance structures for cycloheptatriene anion. Are all bonds equivalent? How many lines (signals) would you see in a H1 NMR? C13 NMR?
Answer
All protons and carbons are the same, so therefore each spectrum will only have one signal each in the proton NMR and carbon NMR.
Exercise \(3\)
The following reaction occurs readily. Propose a reason why this occurs?
Answer
The ring becomes aromatic with the addition of two electrons. Thereby obeying the 4n+2 rule. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/15%3A_Benzene_and_Aromaticity%3A_Electrophilic_Aromatic_Substitution/15.07%3A_Huckel%27s_Rule__and_Charged__Molecules.txt |
Bensene contains six pi electrons which are delocalized in six p orbitals above and below the plane of the benzene ring.
The six pi electrons obey Huckel's rule so benzene is especially stable. This means that the aromatic ring want to be retained during reactions. Because of this benzene does not undergo addition like other unsaturated hydrocarbons.
Benzene can undergo electrophilic aromatic substitution because aromaticity is maintained.
The General Mechanism
Step 1 (Slow)
The e- in the pi bond attacks the electrophile
One carbon gets a positive charge the other forms a C-E bond
This forms the arenium ion.
The arenium ion is conjugated but not aromatic.
Step 2 (Fast)
The LPE on a base attacks the hydrogen.
This causes the e- in the C-H bond to form a C-C double bond and aromaticity is reformed
A Detailed discussion of the Mechanism for Electrophilic Substitution Reactions of Benzene
A two-step mechanism has been proposed for these electrophilic substitution reactions. In the first, slow or rate-determining, step the electrophile forms a sigma-bond to the benzene ring, generating a positively charged benzenonium intermediate. In the second, fast step, a proton is removed from this intermediate, yielding a substituted benzene ring. The following four-part illustration shows this mechanism for the bromination reaction. Also, an animated diagram may be viewed.
Preliminary step: Formation of the strongly electrophilic bromine cation
Step 1: The electrophile forms a sigma-bond to the benzene ring, generating a positively charged benzenonium intermediate
Step 2: A proton is removed from this intermediate, yielding a substituted benzene ring
This mechanism for electrophilic aromatic substitution should be considered in context with other mechanisms involving carbocation intermediates. These include SN1 and E1 reactions of alkyl halides, and Brønsted acid addition reactions of alkenes.
To summarize, when carbocation intermediates are formed one can expect them to react further by one or more of the following modes:
1. The cation may bond to a nucleophile to give a substitution or addition product.
2. The cation may transfer a proton to a base, giving a double bond product.
3. The cation may rearrange to a more stable carbocation, and then react by mode #1 or #2.
SN1 and E1 reactions are respective examples of the first two modes of reaction. The second step of alkene addition reactions proceeds by the first mode, and any of these three reactions may exhibit molecular rearrangement if an initial unstable carbocation is formed. The carbocation intermediate in electrophilic aromatic substitution (the benzenonium ion) is stabilized by charge delocalization (resonance) so it is not subject to rearrangement. In principle it could react by either mode 1 or 2, but the energetic advantage of reforming an aromatic ring leads to exclusive reaction by mode 2 (ie. proton loss).
Other Examples of Electophilic Aromatic Substitution
Many other substitution reactions of benzene have been observed, the five most useful are listed below (chlorination and bromination are the most common halogenation reactions). Since the reagents and conditions employed in these reactions are electrophilic, these reactions are commonly referred to as Electrophilic Aromatic Substitution. The catalysts and co-reagents serve to generate the strong electrophilic species needed to effect the initial step of the substitution. The specific electrophile believed to function in each type of reaction is listed in the right hand column.
Reaction Type Typical Equation Electrophile E(+)
Halogenation: C6H6 + Cl2 & heat
FeCl3 catalyst
——> C6H5Cl + HCl
Chlorobenzene
Cl(+) or Br(+)
Nitration: C6H6 + HNO3 & heat
H2SO4 catalyst
——> C6H5NO2 + H2O
Nitrobenzene
NO2(+)
Sulfonation: C6H6 + H2SO4 + SO3
& heat
——> C6H5SO3H + H2O
Benzenesulfonic acid
SO3H(+)
Alkylation:
Friedel-Crafts
C6H6 + R-Cl & heat
AlCl3 catalyst
——> C6H5-R + HCl
An Arene
R(+)
Acylation:
Friedel-Crafts
C6H6 + RCOCl & heat
AlCl3 catalyst
——> C6H5COR + HCl
An Aryl Ketone
RCO(+)
Contributors
• Prof. Steven Farmer (Sonoma State University)
• William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/15%3A_Benzene_and_Aromaticity%3A_Electrophilic_Aromatic_Substitution/15.08%3A_Synthesis_of_Benzene__Derivatives%3A__Electrophilic_Aromatic__Substitution.txt |
Halogenation is an example of electrophillic aromatic substitution. In electrophilic aromatic substitutions, a benzene is attacked by an electrophile which results in substition of hydrogens. However, halogens are not electrophillic enough to break the aromaticity of benzenes, which require a catalyst to activate.
Activation of Halogen
(where X= Br or Cl, we will discuss further in detail later why other members of the halogen family Flourine and Iodine are not used in halogenation of benzenes)
Hence, Halogen needs the help and aid of Lewis Acidic Catalysts to activate it to become a very strong electrophile. Examples of these activated halogens are Ferric Hallides (FeX3) Aluminum Halides (AlX3) where X= Br or Cl. In the following examples, the halogen we will look at is Bromine.
In the example of bromine, in order to make bromine electrophillic enough to react with benzene, we use the aid of an aluminum halide such as aluminum bromide.
With aluminum bromide as a Lewis acid, we can mix Br2 with AlBr3 to give us Br+. The presence of Br+ is a much better electrophile than Br2 alone. Bromination is acheived with the help of AlBr3 (Lewis acid catalysts) as it polarizes the Br-Br bond. The polarization causes polarization causes the bromine atoms within the Br-Br bond to become more electrophillic. The presence of Br+ compared to Br2 alone is a much better electrophille that can then react with benzene.
As the bromine has now become more electrophillic after activation of a catalyst, an electrophillic attack by the benzene occurs at the terminal bromine of Br-Br-AlBr3. This allows the other bromine atom to leave with the AlBr3 as a good leaving group, AlBr4-.
After the electrophilic attack of bromide to the benzene, the hydrogen on the same carbon as bromine substitutes the carbocation in which resulted from the attack. Hence it being an electrophilic aromatic SUBSTITUTION. Since the by-product aluminum tetrabromide is a strong nucleophile, it pulls of a proton from the Hydrogen on the same carbon as bromine.
In the end, AlBr3was not consumed by the reaction and is regenerated. It serves as our catalyst in the halogenation of benzenes.
Dissociation Energies of Halogens and its Effect on Halogenation of Benzenes
The electrophillic bromination of benzenes is an exothermic reaction. Considering the exothermic rates of aromatic halogenation decreasing down the periodic table in the Halogen family, Flourination is the most exothermic and Iodination would be the least. Being so exothermic, a reaction of flourine with benzene is explosive! For iodine, electrophillic iodination is generally endothermic, hence a reaction is often not possible. Similar to bromide, chlorination would require the aid of an activating presence such as Alumnium Chloride or Ferric Chloride. The mechanism of this reaction is the same as with Bromination of benzene.
Problems
1. What reagents would you need to get the given product?
What reagents would you need to gete given product
2. What product would result from the given reagents?
3. What is the major product given the reagents below?
4. Draw the formatin of Cl+ from AlCl3 and Cl2
5. Draw the mechanism of the reaction between Cl+ and a benzene.
Solutions
1. Cl2 and AlCl3 or Cl2 and FeCl3
2. No Reaction
3.
4.
5.
Contributors
• Catherine Nguyen | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/15%3A_Benzene_and_Aromaticity%3A_Electrophilic_Aromatic_Substitution/15.09%3A_Halogenation_of_Benzene%3A__The__Need_for_a_Catalyst.txt |
Nitration and sulfonation of benzene are two examples of electrophilic aromatic substitution. The nitronium ion (NO2+) and sulfur trioxide (SO3) are the electrophiles and individually react with benzene to give nitrobenzene and benzenesulfonic acid respectively.
Nitration of Benzene
The source of the nitronium ion is through the protonation of nitric acid by sulfuric acid, which causes the loss of a water molecule and formation of a nitronium ion.
Sulfuric Acid Activation of Nitric Acid
The first step in the nitration of benzene is to activate HNO3with sulfuric acid to produce a stronger electrophile, the nitronium ion.
Because the nitronium ion is a good electrophile, it is attacked by benzene to produce Nitrobenzene.
Mechanism
(Resonance forms of the intermediate can be seen in the generalized electrophilic aromatic substitution)
Sulfonation of Benzene
Sulfonation is a reversible reaction that produces benzenesulfonic acid by adding sulfur trioxide and fuming sulfuric acid. The reaction is reversed by adding hot aqueous acid to benzenesulfonic acid to produce benzene.
Mechanism
To produce benzenesulfonic acid from benzene, fuming sulfuric acid and sulfur trioxide are added. Fuming sulfuric acid, also refered to as oleum, is a concentrated solution of dissolved sulfur trioxide in sulfuric acid. The sulfur in sulfur trioxide is electrophilic because the oxygens pull electrons away from it because oxygen is very electronegative. The benzene attacks the sulfur (and subsequent proton transfers occur) to produce benzenesulfonic acid.
Reverse Sulfonation
Sulfonation of benzene is a reversible reaction. Sulfur trioxide readily reacts with water to produce sulfuric acid and heat. Therefore, by adding heat to benzenesulfonic acid in diluted aqueous sulfuric acid the reaction is reversed.
Further Applications of Nitration and Sulfonation
Nitration is used to add nitrogen to a benzene ring, which can be used further in substitution reactions. The nitro group acts as a ring deactivator. Having nitrogen present in a ring is very useful because it can be used as a directing group as well as a masked amino group. The products of aromatic nitrations are very important intermediates in industrial chemistry.
Because sulfonation is a reversible reaction, it can also be used in further substitution reactions in the form of a directing blocking group because it can be easily removed. The sulfonic group blocks the carbon from being attacked by other substituents and after the reaction is completed it can be removed by reverse sulfonation. Benzenesulfonic acids are also used in the synthesis of detergents, dyes, and sulfa drugs. Bezenesulfonyl Chloride is a precursor to sulfonamides, which are used in chemotherapy.
Problems
1. What is/are the required reagent(s)for the following reaction:
2. What is the product of the following reaction:
3. Why is it important that the nitration of benzene by nitric acid occurs in sulfuric acid?
4. Write a detailed mechanism for the sulfonation of benzene, including all resonance forms.
5. Draw an energy diagram for the nitration of benzene. Draw the intermediates, starting materials, and products. Label the transition states. (For questions 1 and 2 see Electrophilic Aromatic Substitution for hints)
For other problems involving Electrophilic Aromatic Substitution and similar reactions see:
• Electrophilic Aromatic Substitution
• Activating and Deactivating Benzene Rings
• Electrophilic Attack on Disubstituted Benzenes
Solutions
1. SO3 and H2SO4 (fuming)
2.
3. Sulfuric acid is needed in order for a good electrophile to form. Sulfuric acid protonates nitric acid to form the nitronium ion (water molecule is lost). The nitronium ion is a very good electrophile and is open to attack by benzene. Without sulfuric acid the reaction would not occur.
4.
5. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/15%3A_Benzene_and_Aromaticity%3A_Electrophilic_Aromatic_Substitution/15.10%3A_Nitration__and_Sulfonation_of_Benzene.txt |
Friedel-Crafts Alkylation
Friedel-Crafts Alkylation was first discovered by French scientist Charles Friedel and his partner, American scientist James Crafts, in 1877. This reaction allowed for the formation of alkyl benzenes from alkyl halides, but was plagued with unwanted supplemental activity that reduced its efficiency.
The mechanism takes place as follows:
Step one:
The first step creates a cabocation that acts as the electrophile in the reaction. This step activates the haloalkane. Secondary and teriary halides only form the free cabocation in the step.
Step two
The second step has an electrophilic attack on the benzene that results in multiple resonance forms. The halogen reactions with the intermediate and picks up the hydrogen to eliminate the positive charge.
Finish
The final step shown above is the results of the end of step and shows the final products.
The reactivity of haloalkanes increases as you move up the periodic table and increase polarity. This means that an RF haloalkane is most reactive followed by RCl then RBr and finally RI. This means that the Lewis acids used as catalysts in Friedel-Crafts Alkylation reactions tend have similar halogen combinations such as BF3, SbCl5, AlCl3, SbCl5, and AlBr3, all of which are commonly used in these reactions.
Some limitations of Friedel-Crafts Alkylation
There are possibilities of carbocation rearrangements when you are trying to add a carbon chain greater than two carbons. The rearrangements occur due to hydride shifts and methyl shifts. For example, the product of a Friedel-Crafts Alkylation will show an iso rearrangement when adding a three carbon chain as a substituent. Also, the reaction will only work if the ring you are adding a substituent to is not deactivated. For a look at substituents that activating or deactivating Benzene Rings.
The three key limitations of Friedel-Crafts alkylation are:
1. Carbocation Rearrangement - Only certain alkylbenzenes can be made due to the tendency of cations to rearrange.
2. Compound Limitations - Friedel-Crafts fails when used with compounds such as nitrobenzene and other strong deactivating systems.
3. Polyalkylation - Products of Friedel-Crafts are even more reactive than starting material. Alkyl groups produced in Friedel-Crafts Alkylation are electron-donating substituents meaning that the products are more susceptible to electrophilic attack than what we began with. For synthetic purposes, this is a big dissapointment.
To remedy these limitations, a new and improved reaction was devised: The Friedel-Crafts Acylation. (also known as Friedel-Crafts Alkanoylation).
Friedel-Crafts Acylation
The goal of the reaction is the following:
The very first step involves the formation of the acylium ion which will later react with benzene:
The second step involves the attack of the acylium ion on benzene as a new electrophile to form one complex:
The third step involves the departure of the proton in order for aromaticity to return to benzene:
During the third step, AlCl4 returns to remove a proton from the benzene ring, which enables the ring to return to aromaticity. In doing so, the original AlCl3 is regenerated for use again, along with HCl. Most importantly, we have the first part of the final product of the reaction, which is a ketone. Thie first part of the product is the complex with aluminum chloride as shown:
The final step involves the addition of water to liberate the final product as the acylbenzene:
Because the acylium ion (as was shown in step one) is stabilized by resonance, no rearrangement occurs (Limitation 1). Also, because of of the deactivation of the product, it is no longer susceptible to electrophilic attack and hence, is no longer susceptible to electrophilic attack and hence, no longer goes into further reactions (Limitation 3). However, as not all is perfect, Limitation 2 still prevails where Friedel-Crafts Acylation fails with strong deactivating rings. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/15%3A_Benzene_and_Aromaticity%3A_Electrophilic_Aromatic_Substitution/15.11%3A_%09Friedel-Crafts_Alkylation.txt |
Friedel-Crafts Alkylation
Friedel-Crafts Alkylation was first discovered by French scientist Charles Friedel and his partner, American scientist James Crafts, in 1877. This reaction allowed for the formation of alkyl benzenes from alkyl halides, but was plagued with unwanted supplemental activity that reduced its efficiency.
The mechanism takes place as follows:
Step one:
The first step creates a cabocation that acts as the electrophile in the reaction. This step activates the haloalkane. Secondary and teriary halides only form the free cabocation in the step.
Step two
The second step has an electrophilic attack on the benzene that results in multiple resonance forms. The halogen reactions with the intermediate and picks up the hydrogen to eliminate the positive charge.
Finish
The final step shown above is the results of the end of step and shows the final products.
The reactivity of haloalkanes increases as you move up the periodic table and increase polarity. This means that an RF haloalkane is most reactive followed by RCl then RBr and finally RI. This means that the Lewis acids used as catalysts in Friedel-Crafts Alkylation reactions tend have similar halogen combinations such as BF3, SbCl5, AlCl3, SbCl5, and AlBr3, all of which are commonly used in these reactions.
Some limitations of Friedel-Crafts Alkylation
There are possibilities of carbocation rearrangements when you are trying to add a carbon chain greater than two carbons. The rearrangements occur due to hydride shifts and methyl shifts. For example, the product of a Friedel-Crafts Alkylation will show an iso rearrangement when adding a three carbon chain as a substituent. Also, the reaction will only work if the ring you are adding a substituent to is not deactivated. For a look at substituents that activating or deactivating Benzene Rings.
The three key limitations of Friedel-Crafts alkylation are:
1. Carbocation Rearrangement - Only certain alkylbenzenes can be made due to the tendency of cations to rearrange.
2. Compound Limitations - Friedel-Crafts fails when used with compounds such as nitrobenzene and other strong deactivating systems.
3. Polyalkylation - Products of Friedel-Crafts are even more reactive than starting material. Alkyl groups produced in Friedel-Crafts Alkylation are electron-donating substituents meaning that the products are more susceptible to electrophilic attack than what we began with. For synthetic purposes, this is a big dissapointment.
To remedy these limitations, a new and improved reaction was devised: The Friedel-Crafts Acylation. (also known as Friedel-Crafts Alkanoylation).
Friedel-Crafts Acylation
The goal of the reaction is the following:
The very first step involves the formation of the acylium ion which will later react with benzene:
The second step involves the attack of the acylium ion on benzene as a new electrophile to form one complex:
The third step involves the departure of the proton in order for aromaticity to return to benzene:
During the third step, AlCl4 returns to remove a proton from the benzene ring, which enables the ring to return to aromaticity. In doing so, the original AlCl3 is regenerated for use again, along with HCl. Most importantly, we have the first part of the final product of the reaction, which is a ketone. Thie first part of the product is the complex with aluminum chloride as shown:
The final step involves the addition of water to liberate the final product as the acylbenzene:
Because the acylium ion (as was shown in step one) is stabilized by resonance, no rearrangement occurs (Limitation 1). Also, because of of the deactivation of the product, it is no longer susceptible to electrophilic attack and hence, is no longer susceptible to electrophilic attack and hence, no longer goes into further reactions (Limitation 3). However, as not all is perfect, Limitation 2 still prevails where Friedel-Crafts Acylation fails with strong deactivating rings. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/15%3A_Benzene_and_Aromaticity%3A_Electrophilic_Aromatic_Substitution/15.12%3A_Limitations_of_Friedel-Crafts_Alkylations.txt |
Friedel-Crafts Alkylation
Friedel-Crafts Alkylation was first discovered by French scientist Charles Friedel and his partner, American scientist James Crafts, in 1877. This reaction allowed for the formation of alkyl benzenes from alkyl halides, but was plagued with unwanted supplemental activity that reduced its efficiency.
The mechanism takes place as follows:
Step one:
The first step creates a cabocation that acts as the electrophile in the reaction. This step activates the haloalkane. Secondary and teriary halides only form the free cabocation in the step.
Step two
The second step has an electrophilic attack on the benzene that results in multiple resonance forms. The halogen reactions with the intermediate and picks up the hydrogen to eliminate the positive charge.
Finish
The final step shown above is the results of the end of step and shows the final products.
The reactivity of haloalkanes increases as you move up the periodic table and increase polarity. This means that an RF haloalkane is most reactive followed by RCl then RBr and finally RI. This means that the Lewis acids used as catalysts in Friedel-Crafts Alkylation reactions tend have similar halogen combinations such as BF3, SbCl5, AlCl3, SbCl5, and AlBr3, all of which are commonly used in these reactions.
Some limitations of Friedel-Crafts Alkylation
There are possibilities of carbocation rearrangements when you are trying to add a carbon chain greater than two carbons. The rearrangements occur due to hydride shifts and methyl shifts. For example, the product of a Friedel-Crafts Alkylation will show an iso rearrangement when adding a three carbon chain as a substituent. Also, the reaction will only work if the ring you are adding a substituent to is not deactivated. For a look at substituents that activating or deactivating Benzene Rings.
The three key limitations of Friedel-Crafts alkylation are:
1. Carbocation Rearrangement - Only certain alkylbenzenes can be made due to the tendency of cations to rearrange.
2. Compound Limitations - Friedel-Crafts fails when used with compounds such as nitrobenzene and other strong deactivating systems.
3. Polyalkylation - Products of Friedel-Crafts are even more reactive than starting material. Alkyl groups produced in Friedel-Crafts Alkylation are electron-donating substituents meaning that the products are more susceptible to electrophilic attack than what we began with. For synthetic purposes, this is a big dissapointment.
To remedy these limitations, a new and improved reaction was devised: The Friedel-Crafts Acylation. (also known as Friedel-Crafts Alkanoylation).
Friedel-Crafts Acylation
The goal of the reaction is the following:
The very first step involves the formation of the acylium ion which will later react with benzene:
The second step involves the attack of the acylium ion on benzene as a new electrophile to form one complex:
The third step involves the departure of the proton in order for aromaticity to return to benzene:
During the third step, AlCl4 returns to remove a proton from the benzene ring, which enables the ring to return to aromaticity. In doing so, the original AlCl3 is regenerated for use again, along with HCl. Most importantly, we have the first part of the final product of the reaction, which is a ketone. Thie first part of the product is the complex with aluminum chloride as shown:
The final step involves the addition of water to liberate the final product as the acylbenzene:
Because the acylium ion (as was shown in step one) is stabilized by resonance, no rearrangement occurs (Limitation 1). Also, because of of the deactivation of the product, it is no longer susceptible to electrophilic attack and hence, is no longer susceptible to electrophilic attack and hence, no longer goes into further reactions (Limitation 3). However, as not all is perfect, Limitation 2 still prevails where Friedel-Crafts Acylation fails with strong deactivating rings. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/15%3A_Benzene_and_Aromaticity%3A_Electrophilic_Aromatic_Substitution/15.13%3A_Friedel-Crafts_Alkanoylation_%28Acylation%29.txt |
Objectives
After completing this section, you should be able to
1. describe the two ways in which a substituent influences the electrophilic substitution of a monosubstituted aromatic compound.
2. classify each of the following substituents as being either activating or deactivating with respect to electrophilic aromatic substitution: \(\ce{\sf{-NH2}}\), \(\ce{\sf{-OH}}\), \(\ce{\sf{-NHR}}\), \(\ce{\sf{-NR2}}\), \(\ce{\sf{-OR}}\), \(\ce{\sf{-NHCOR}}\), alkyl (R), phenyl, \(\ce{\sf{R3N+}}\), \(\ce{\sf{-NO2}}\), \(\ce{\sf{-CN}}\), \(\ce{\sf{-COR}}\), \(\ce{\sf{-CO2H}}\), \(\ce{\sf{-CO2R}}\), \(\ce{\sf{-CHO}}\), halogens.
3. list a given series of substituents (selected from those given in Objective 2) in order of increasing or decreasing ability to activate or deactivate an aromatic ring with respect to electrophilic substitution.
4. explain, in general terms, the factors that determine whether a given substituent will activate or deactivate an aromatic ring with respect to electrophilic substitution.
5. list a given series of aromatic compounds in order of increasing or decreasing reactivity with respect to electrophilic substitution.
6. explain the inductive effects displayed by substituents such as nitro, carboxyl, alkyl and the halogens during electrophilic aromatic substitution reactions.
7. explain the resonance effects displayed by substituents such as nitro, carbonyl-containing, hydroxy, alkoxy and amino groups during electrophilic aromatic substitution reactions.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• inductive effect
• resonance effect
Study Notes
On reading Objective 2 students may exclaim “How am I ever going to memorize all of this!”—or words to that effect. The answer is that if you are trying to memorize such things, you are taking the wrong approach to organic chemistry. What you should be doing is trying to understand the factors that determine whether a given substituent will activate or deactivate a benzene ring with respect to electrophilic substitution.
You may wish to review earlier material on to the inductive effect. If so, refer to Sections 2.1, 7.9 (paying particular attention to the “Study Notes”) and 14.5.
Note that one argument sometimes used to explain the ability of alkyl groups to donate electrons inductively to an aromatic ring is that sp2‑hybridized carbon atoms are more electronegative than sp3‑hybridized carbon atoms. Thus, a sigma bond between sp2- and sp3‑carbon is slightly polarized, as follows:
When substituted benzene compounds undergo electrophilic substitution reactions of the kind discussed above, two related features must be considered:
I. The first is the relative reactivity of the compound compared with benzene itself. Experiments have shown that substituents on a benzene ring can influence reactivity in a profound manner. For example, a hydroxy or methoxy substituent increases the rate of electrophilic substitution about ten thousand fold, as illustrated by the case of anisole in the virtual demonstration (above). In contrast, a nitro substituent decreases the ring's reactivity by roughly a million. This activation or deactivation of the benzene ring toward electrophilic substitution may be correlated with the electron donating or electron withdrawing influence of the substituents, as measured by molecular dipole moments. In the following diagram we see that electron donating substituents (blue dipoles) activate the benzene ring toward electrophilic attack, and electron withdrawing substituents (red dipoles) deactivate the ring (make it less reactive to electrophilic attack).
The influence a substituent exerts on the reactivity of a benzene ring may be explained by the interaction of two effects:
The first is the inductive effect of the substituent. Most elements other than metals and carbon have a significantly greater electronegativity than hydrogen. Consequently, substituents in which nitrogen, oxygen and halogen atoms form sigma-bonds to the aromatic ring exert an inductive electron withdrawal, which deactivates the ring (left-hand diagram below).
The second effect is the result of conjugation of a substituent function with the aromatic ring. This conjugative interaction facilitates electron pair donation or withdrawal, to or from the benzene ring, in a manner different from the inductive shift. If the atom bonded to the ring has one or more non-bonding valence shell electron pairs, as do nitrogen, oxygen and the halogens, electrons may flow into the aromatic ring by p-π conjugation (resonance), as in the middle diagram. Finally, polar double and triple bonds conjugated with the benzene ring may withdraw electrons, as in the right-hand diagram. Note that in the resonance examples all the contributors are not shown. In both cases the charge distribution in the benzene ring is greatest at sites ortho and para to the substituent.
In the case of the nitrogen and oxygen activating groups displayed in the top row of the previous diagram, electron donation by resonance dominates the inductive effect and these compounds show exceptional reactivity in electrophilic substitution reactions. Although halogen atoms have non-bonding valence electron pairs that participate in p-π conjugation, their strong inductive effect predominates, and compounds such as chlorobenzene are less reactive than benzene. The three examples on the left of the bottom row (in the same diagram) are examples of electron withdrawal by conjugation to polar double or triple bonds, and in these cases the inductive effect further enhances the deactivation of the benzene ring. Alkyl substituents such as methyl increase the nucleophilicity of aromatic rings in the same fashion as they act on double bonds.
Exercises
Exercise \(1\)
Draw the resonance structures for benzaldehyde to show the electron-withdrawing group.
Answer
Exercise \(2\)
Draw the resonance structures for methoxybenzene to show the electron-donating group.
Answer | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/16%3A_Electrophilic_Attack_on_Derivatives_of_Benzene%3A_Substituents_Control_Regioselectivity/16.1%3A_Activation__or__Deactivation__by_Substituents_on_a_Ben.txt |
Objectives
After completing this section, you should be able to
1. draw the resonance contributors for the carbocation intermediate formed during the reaction of a given monosubstituted benzene derivative with any of the electrophiles discussed in this chapter.
2. classify each of the substituents listed in Objective 2 of Section 16.4 as being either meta or ortho/para directing.
3. classify each of the substituents listed in Objective 2 of Section 16.4 as being ortho/para directing activators, ortho/para directing deactivators, or meta directing deactivators.
4. predict the product or products formed from the reaction of a given monosubstituted benzene derivative with each of the electrophiles discussed in this chapter.
5. explain, by drawing the resonance contributors for the intermediate carbocation, why the electrophilic substitution of an alkyl benzene results in a mixture of mainly ortho- and para- substituted products.
6. explain why the electrophilic substitution of phenols, amines and their derivatives proceeds more rapidly than the electrophilic substitution of benzene itself.
7. explain, by drawing the resonance contributors for the intermediate carbocation, why meta substitution predominates in electrophilic aromatic substitution reactions carried out on benzene derivatives containing one of the substituents R3N+, NO2, CO2H, CN, CO2R, COR or CHO.
8. explain why electrophilic aromatic substitution of benzene derivatives containing one of the substituents listed in Objective 7, above, proceeds more slowly than the electrophilic substitution of benzene itself.
9. explain, by drawing the resonance contributors for the intermediate carbocation, why the electrophilic aromatic substitution of halobenzenes produces a mixture of mainly ortho- and para-substituted products.
10. explain why the electrophilic aromatic substitution of halobenzenes proceeds more slowly than does the electrophilic substitution of benzene itself.
11. use the principles developed in this chapter to predict in which of the three categories listed in Objective 3, above, a previously unencountered substituent should be placed.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• steric effect
• steric hindrance
Study Notes
As you saw in Section 16.4, a substituent on a benzene ring can be an activator or a deactivator. At the same time, a substituent can also be a meta director or an ortho/para director. Of the four possible combinations, only three are known—there are no meta directing activators.
If you look at the data for the nitration of toluene, you will see that the yield of o‑nitrotoluene is 63% and that of p‑nitrotoluene is 34%. Statistically, we should expect to obtain twice as much ortho product as para product, because the former is produced by attack at either of two carbon atoms whereas the latter is produced by attack at only one carbon atom (see Figure below).
In this instance, the observed ortho/para ratio is almost 2:1, as we might expect. However, if we study the ortho/para ratio found in the nitration of a number of other arenes, we see that this is not always the case. Note that the data for the nitration of toluene given in the table below differ from those presented elsewhere. The variation may result from a difference in temperature, reaction conditions or reagent, and emphasizes the point that it is the trends which are important, not the numbers themselves.
Substrate % ortho % para ortho/para ratio
toluene 58 37 1.57:1
ethylbenzene 45 49 0.92:1
isopropylbenzene 30 62 0.48:1
tert-butylbenzene 16 73 0.22:1
[Source: These data were taken from the audiocassette Some Organic Reaction Pathways, by Peter Sykes. London: Educational Techniques Subject Group, The Chemical Society, 1975.]
Table $1$: Nitration of arenes
The table above shows us that as the size of the alkyl substituent already present in the ring increases, attack at the ortho position becomes more difficult, and the percentage of ortho isomers in the mixture of products decreases. This is an example of a steric effect—an effect caused by the size of the substituent—and we would say that as the size of the alkyl group increases, attack at the ortho position becomes less favorable as a result of steric hindrance. Note that the size of the electrophile can also be a factor in determining the ortho/para ratio: the larger the electrophile, the less able it is to attack at the ortho position, particularly if the substituent already present in the ring is itself quite bulky.
When drawing the resonance contributors to the carbocation formed during an electrophilic aromatic substitution, bear in mind that those of the type
are particularly important, because in such structures each atom possesses a complete octet of electrons.
Note that, as do the hydroxyl and amino groups, the halogens have an inductive electron-withdrawing effect and a resonance electron-releasing effect on a benzene ring. The difference in behavior during electrophilic substitutions arises because, with the hydroxyl and amino groups, the resonance effect is much greater than the inductive effect, whereas with the halogens, there is a much finer balance. In the case of the latter, the inductive effect reduces the overall reactivity, but the resonance effect means that this reduction is felt less at the ortho and para positions than at the meta position.
Substituted rings are divided into two groups based on the type of the substituent that the ring carries:
• Activated rings: the substituents on the ring are groups that donate electrons.
• Deactivated rings: the substituents on the ring are groups that withdraw electrons.
Introduction
Examples of activating groups in the relative order from the most activating group to the least activating:
-NH2, -NR2 > -OH, -OR> -NHCOR> -CH3 and other alkyl groups
with R as alkyl groups (CnH2n+1)
Examples of deactivating groups in the relative order from the most deactivating to the least deactivating:
-NO2, -CF3> -COR, -CN, -CO2R, -SO3H > Halogens
with R as alkyl groups (CnH2n+1)
The order of reactivity among Halogens from the more reactive (least deactivating substituent) to the least reactive (most deactivating substituent) halogen is:
F> Cl > Br > I
The order of reactivity of the benzene rings toward the electrophilic substitution when it is substituted with a halogen groups, follows the order of electronegativity. The ring that is substituted with the most electronegative halogen is the most reactive ring (less deactivating substituent) and the ring that is substituted with the least electronegative halogen is the least reactive ring (more deactivating substituent), when we compare rings with halogen substituents. Also the size of the halogen effects the reactivity of the benzene ring that the halogen is attached to. As the size of the halogen increase, the reactivity of the ring decreases.
The direction of the reaction
The activating group directs the reaction to the ortho or para position, which means the electrophile substitutes for the hydrogen that is on carbon 2 or carbon 4. The deactivating group directs the reaction to the meta position, which means the electrophile substitutes for the hydrogen that is on carbon 3 with the exception of the halogens which are deactivating groups but direct the ortho or para substitution.
Substituents determine the reaction direction by resonance or inductive effect
Resonance effect is the conjugation between the ring and the substituent, which means the delocalizing of the $\pi$ electrons between the ring and the substituent. Inductive effect is the withdraw of the sigma ( the single bond ) electrons away from the ring toward the substituent, due to the higher electronegativity of the substituent compared to the carbon of the ring.
Activating groups (ortho or para directors)
When substituents such as -OH have an unshared pair of electrons, the resonance effect is stronger than the inductive effect which make these substituents stronger activators, since this resonance effect direct the electron toward the ring. In cases where the subtituents is esters or amides, they are less activating because they form resonance structure that pull the electron density away from the ring.
By looking at the mechanism above, we can see how electron donating groups direct electrophilic substitution to the ortho and para positions. Since the extra electron density is localized on the ortho and para carbons, these carbons are more likely to react with the electrophile.
Inductive effects of alkyl groups activate the direction of the ortho or para substitution, which is when s electrons gets pushed toward the ring.
Deactivating group (meta directors)
The deactivating groups deactivate the ring by the inductive effect in the presence of an electronegative atom that withdraws electron density away from the ring.
The mechanism above shows that when electron density is withdrawn from the ring, that leaves the carbons at the ortho, para positions with a parital positive charge which is unfavorable for the electrophile, so the electrophile attacks the carbon at the meta positions.
Halogens are an exception of the deactivating group that directs to the ortho or para substitution. The halogens deactivate the ring by inductive effect not by the resonance even though they have an unpaired pair of electrons. The unpaired pair of electrons gets donated to the ring, but the inductive effect pulls away the s electrons from the ring by the electronegativity of the halogens.
Substituents determine the reactivity of rings
The reaction of a substituted ring with an activating group is faster than the same reaction wtih benzene. On the other hand, a substituted ring with a deactivated group reacts slower than benzene.
Activating groups speed up reaction with electrophiles due to increased electron density on the ring. This stabilizes the intermediate carbocation, which decreases the activation energy for the reaction. On the other hand, deactivating groups withdraw electron density away from the carbocation formed in the intermediate step, increasing the activation energy, which slows down the reaction.
The CH3 Group is an ortho, para director
Alkyl groups are inductively donating, therefore are activators. This resulsts in o/p attack to form a tertiary arenium carbocation which speeds up the reaction.
The O-CH3 Group is an ortho, para director
The methoxy group is an example of groups that are ortho, para directors by having and oxygen or nitrogen adjacent to the aromatic ring. This same activation is present with alcohols, amines, esters and amides (with the oxygen or nitrogen attached to the ring, not the carbonyl).
Groups with an oxygen or nitrogen attached to the aromatic ring are ortho and para directors since the O or N can push electrons into the ring, making the ortho and para positions more reactive and stabilizing the arenium ion that forms. This causes the ortho and para products to form faster than meta. Generally, the para product is preferred because of steric effects.
Acyl groups are meta directors
Ketones are an example of groups that deactivate an aromatic ring through resonance. Similar deactivation also occurs with ammonium ions, nitro groups, aldehydes, nitriles, sulfonic acids, and groups with a carbonyl attached to the ring (amides, esters, carboxylic acids, and anhydrides).
Acyl groups are resonance deactivators. Ortho and para attack produces a resonance structure which places the arenium cation next to an additional cation. This destabilizes the arenium cation and slows down ortho and para reaction. By default the meta product forms faster because it lacks this destabilizing resonance structure.
Halogens
Halogens are an interesting hybrid case. They are ortho, para directors, but deactivators. Overall, they remove electron density from the ring, making it less reactive. However, due to their resonance donation to the ring, if it does react, it reacts primarily at ortho and para positions.
References
1. Schore, N.E. and P.C. Vollhardt. 2007. Organic Chemistry, structure and function, 5th ed. New York,NY: W.H. Freeman and Company.
2. Fryhle, C.B. and G. Solomons. 2008. Organic Chemistry, 9th ed.Danvers,MA: Wiley.
Exercise $1$
Predict the pattern of the electrophilic substitution on these rings:
Answer
The first substitution is going to be ortho and/or para substitution since we have a halogen subtituent. The second substitution is going to be ortho and/or para substitution also since we have an alkyl substituent.
Exercise $2$
Which nitration product is going to form faster: nitration of aniline or nitration of nitrobenzene?
Answer
The nitration of aniline is going to be faster than the nitration of nitrobenzene, since the aniline is a ring with NH2 substituent and nitrobenzene is a ring with NO2 substituent. As described above NH2 is an activating group which speeds up the reaction and NO2 is a deactivating group that slows down the reaction.
Exercise $3$
Predict the product of the following sulfonation reaction:
Answer
Exercise $4$
Classify these two groups as activating or deactivating groups:
A. alcohol
B. ester
Answer
A. alcohol is an activating group.
B. Esters can be either. If the oxygen atom is next to the ring, esters are activating. However if the carbonyl is next to the ring, the ester is a deactivating group.
Exercise $5$
Does a chloride substituent activate or deactivate an aromatic ring?
Answer
Chloride deactivate an aromatic ring due to the inductive effect.
Exercise $6$
(Trichloromethyl)benzene has a strong concentration of electrons at the methyl substituent. Comparing this toluene, which is more reactive toward electrophilic substitution?
Answer
The trichloromethyl group is an electron donor into the benzene ring, therefore making it more stable and therefore more reactive compared to electrophilic substitution.
Exercise $7$
The following compound is less reactive towards electrophilic substitution than aniline? Explain.
Answer
As seen in resonance the electron density is also localized off of the ring, thereby deactivating it compared to aniline.
Exercise $8$
Consider the intermediates of the following molecule during an electrophilic substitution. Draw resonance structures for ortho, meta, and para attacks.
Answer | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/16%3A_Electrophilic_Attack_on_Derivatives_of_Benzene%3A_Substituents_Control_Regioselectivity/16.2%3A_Directing_Inductive__Effects__of_Alkyl__Groups.txt |
Objectives
After completing this section, you should be able to
1. draw the resonance contributors for the carbocation intermediate formed during the reaction of a given monosubstituted benzene derivative with any of the electrophiles discussed in this chapter.
2. classify each of the substituents listed in Objective 2 of Section 16.4 as being either meta or ortho/para directing.
3. classify each of the substituents listed in Objective 2 of Section 16.4 as being ortho/para directing activators, ortho/para directing deactivators, or meta directing deactivators.
4. predict the product or products formed from the reaction of a given monosubstituted benzene derivative with each of the electrophiles discussed in this chapter.
5. explain, by drawing the resonance contributors for the intermediate carbocation, why the electrophilic substitution of an alkyl benzene results in a mixture of mainly ortho- and para- substituted products.
6. explain why the electrophilic substitution of phenols, amines and their derivatives proceeds more rapidly than the electrophilic substitution of benzene itself.
7. explain, by drawing the resonance contributors for the intermediate carbocation, why meta substitution predominates in electrophilic aromatic substitution reactions carried out on benzene derivatives containing one of the substituents R3N+, NO2, CO2H, CN, CO2R, COR or CHO.
8. explain why electrophilic aromatic substitution of benzene derivatives containing one of the substituents listed in Objective 7, above, proceeds more slowly than the electrophilic substitution of benzene itself.
9. explain, by drawing the resonance contributors for the intermediate carbocation, why the electrophilic aromatic substitution of halobenzenes produces a mixture of mainly ortho- and para-substituted products.
10. explain why the electrophilic aromatic substitution of halobenzenes proceeds more slowly than does the electrophilic substitution of benzene itself.
11. use the principles developed in this chapter to predict in which of the three categories listed in Objective 3, above, a previously unencountered substituent should be placed.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• steric effect
• steric hindrance
Study Notes
As you saw in Section 16.4, a substituent on a benzene ring can be an activator or a deactivator. At the same time, a substituent can also be a meta director or an ortho/para director. Of the four possible combinations, only three are known—there are no meta directing activators.
If you look at the data for the nitration of toluene, you will see that the yield of o‑nitrotoluene is 63% and that of p‑nitrotoluene is 34%. Statistically, we should expect to obtain twice as much ortho product as para product, because the former is produced by attack at either of two carbon atoms whereas the latter is produced by attack at only one carbon atom (see Figure below).
In this instance, the observed ortho/para ratio is almost 2:1, as we might expect. However, if we study the ortho/para ratio found in the nitration of a number of other arenes, we see that this is not always the case. Note that the data for the nitration of toluene given in the table below differ from those presented elsewhere. The variation may result from a difference in temperature, reaction conditions or reagent, and emphasizes the point that it is the trends which are important, not the numbers themselves.
Substrate % ortho % para ortho/para ratio
toluene 58 37 1.57:1
ethylbenzene 45 49 0.92:1
isopropylbenzene 30 62 0.48:1
tert-butylbenzene 16 73 0.22:1
[Source: These data were taken from the audiocassette Some Organic Reaction Pathways, by Peter Sykes. London: Educational Techniques Subject Group, The Chemical Society, 1975.]
Table $1$: Nitration of arenes
The table above shows us that as the size of the alkyl substituent already present in the ring increases, attack at the ortho position becomes more difficult, and the percentage of ortho isomers in the mixture of products decreases. This is an example of a steric effect—an effect caused by the size of the substituent—and we would say that as the size of the alkyl group increases, attack at the ortho position becomes less favorable as a result of steric hindrance. Note that the size of the electrophile can also be a factor in determining the ortho/para ratio: the larger the electrophile, the less able it is to attack at the ortho position, particularly if the substituent already present in the ring is itself quite bulky.
When drawing the resonance contributors to the carbocation formed during an electrophilic aromatic substitution, bear in mind that those of the type
are particularly important, because in such structures each atom possesses a complete octet of electrons.
Note that, as do the hydroxyl and amino groups, the halogens have an inductive electron-withdrawing effect and a resonance electron-releasing effect on a benzene ring. The difference in behavior during electrophilic substitutions arises because, with the hydroxyl and amino groups, the resonance effect is much greater than the inductive effect, whereas with the halogens, there is a much finer balance. In the case of the latter, the inductive effect reduces the overall reactivity, but the resonance effect means that this reduction is felt less at the ortho and para positions than at the meta position.
Substituted rings are divided into two groups based on the type of the substituent that the ring carries:
• Activated rings: the substituents on the ring are groups that donate electrons.
• Deactivated rings: the substituents on the ring are groups that withdraw electrons.
Introduction
Examples of activating groups in the relative order from the most activating group to the least activating:
-NH2, -NR2 > -OH, -OR> -NHCOR> -CH3 and other alkyl groups
with R as alkyl groups (CnH2n+1)
Examples of deactivating groups in the relative order from the most deactivating to the least deactivating:
-NO2, -CF3> -COR, -CN, -CO2R, -SO3H > Halogens
with R as alkyl groups (CnH2n+1)
The order of reactivity among Halogens from the more reactive (least deactivating substituent) to the least reactive (most deactivating substituent) halogen is:
F> Cl > Br > I
The order of reactivity of the benzene rings toward the electrophilic substitution when it is substituted with a halogen groups, follows the order of electronegativity. The ring that is substituted with the most electronegative halogen is the most reactive ring (less deactivating substituent) and the ring that is substituted with the least electronegative halogen is the least reactive ring (more deactivating substituent), when we compare rings with halogen substituents. Also the size of the halogen effects the reactivity of the benzene ring that the halogen is attached to. As the size of the halogen increase, the reactivity of the ring decreases.
The direction of the reaction
The activating group directs the reaction to the ortho or para position, which means the electrophile substitutes for the hydrogen that is on carbon 2 or carbon 4. The deactivating group directs the reaction to the meta position, which means the electrophile substitutes for the hydrogen that is on carbon 3 with the exception of the halogens which are deactivating groups but direct the ortho or para substitution.
Substituents determine the reaction direction by resonance or inductive effect
Resonance effect is the conjugation between the ring and the substituent, which means the delocalizing of the $\pi$ electrons between the ring and the substituent. Inductive effect is the withdraw of the sigma ( the single bond ) electrons away from the ring toward the substituent, due to the higher electronegativity of the substituent compared to the carbon of the ring.
Activating groups (ortho or para directors)
When substituents such as -OH have an unshared pair of electrons, the resonance effect is stronger than the inductive effect which make these substituents stronger activators, since this resonance effect direct the electron toward the ring. In cases where the subtituents is esters or amides, they are less activating because they form resonance structure that pull the electron density away from the ring.
By looking at the mechanism above, we can see how electron donating groups direct electrophilic substitution to the ortho and para positions. Since the extra electron density is localized on the ortho and para carbons, these carbons are more likely to react with the electrophile.
Inductive effects of alkyl groups activate the direction of the ortho or para substitution, which is when s electrons gets pushed toward the ring.
Deactivating group (meta directors)
The deactivating groups deactivate the ring by the inductive effect in the presence of an electronegative atom that withdraws electron density away from the ring.
The mechanism above shows that when electron density is withdrawn from the ring, that leaves the carbons at the ortho, para positions with a parital positive charge which is unfavorable for the electrophile, so the electrophile attacks the carbon at the meta positions.
Halogens are an exception of the deactivating group that directs to the ortho or para substitution. The halogens deactivate the ring by inductive effect not by the resonance even though they have an unpaired pair of electrons. The unpaired pair of electrons gets donated to the ring, but the inductive effect pulls away the s electrons from the ring by the electronegativity of the halogens.
Substituents determine the reactivity of rings
The reaction of a substituted ring with an activating group is faster than the same reaction wtih benzene. On the other hand, a substituted ring with a deactivated group reacts slower than benzene.
Activating groups speed up reaction with electrophiles due to increased electron density on the ring. This stabilizes the intermediate carbocation, which decreases the activation energy for the reaction. On the other hand, deactivating groups withdraw electron density away from the carbocation formed in the intermediate step, increasing the activation energy, which slows down the reaction.
The CH3 Group is an ortho, para director
Alkyl groups are inductively donating, therefore are activators. This resulsts in o/p attack to form a tertiary arenium carbocation which speeds up the reaction.
The O-CH3 Group is an ortho, para director
The methoxy group is an example of groups that are ortho, para directors by having and oxygen or nitrogen adjacent to the aromatic ring. This same activation is present with alcohols, amines, esters and amides (with the oxygen or nitrogen attached to the ring, not the carbonyl).
Groups with an oxygen or nitrogen attached to the aromatic ring are ortho and para directors since the O or N can push electrons into the ring, making the ortho and para positions more reactive and stabilizing the arenium ion that forms. This causes the ortho and para products to form faster than meta. Generally, the para product is preferred because of steric effects.
Acyl groups are meta directors
Ketones are an example of groups that deactivate an aromatic ring through resonance. Similar deactivation also occurs with ammonium ions, nitro groups, aldehydes, nitriles, sulfonic acids, and groups with a carbonyl attached to the ring (amides, esters, carboxylic acids, and anhydrides).
Acyl groups are resonance deactivators. Ortho and para attack produces a resonance structure which places the arenium cation next to an additional cation. This destabilizes the arenium cation and slows down ortho and para reaction. By default the meta product forms faster because it lacks this destabilizing resonance structure.
Halogens
Halogens are an interesting hybrid case. They are ortho, para directors, but deactivators. Overall, they remove electron density from the ring, making it less reactive. However, due to their resonance donation to the ring, if it does react, it reacts primarily at ortho and para positions.
References
1. Schore, N.E. and P.C. Vollhardt. 2007. Organic Chemistry, structure and function, 5th ed. New York,NY: W.H. Freeman and Company.
2. Fryhle, C.B. and G. Solomons. 2008. Organic Chemistry, 9th ed.Danvers,MA: Wiley.
Exercise $1$
Predict the pattern of the electrophilic substitution on these rings:
Answer
The first substitution is going to be ortho and/or para substitution since we have a halogen subtituent. The second substitution is going to be ortho and/or para substitution also since we have an alkyl substituent.
Exercise $2$
Which nitration product is going to form faster: nitration of aniline or nitration of nitrobenzene?
Answer
The nitration of aniline is going to be faster than the nitration of nitrobenzene, since the aniline is a ring with NH2 substituent and nitrobenzene is a ring with NO2 substituent. As described above NH2 is an activating group which speeds up the reaction and NO2 is a deactivating group that slows down the reaction.
Exercise $3$
Predict the product of the following sulfonation reaction:
Answer
Exercise $4$
Classify these two groups as activating or deactivating groups:
A. alcohol
B. ester
Answer
A. alcohol is an activating group.
B. Esters can be either. If the oxygen atom is next to the ring, esters are activating. However if the carbonyl is next to the ring, the ester is a deactivating group.
Exercise $5$
Does a chloride substituent activate or deactivate an aromatic ring?
Answer
Chloride deactivate an aromatic ring due to the inductive effect.
Exercise $6$
(Trichloromethyl)benzene has a strong concentration of electrons at the methyl substituent. Comparing this toluene, which is more reactive toward electrophilic substitution?
Answer
The trichloromethyl group is an electron donor into the benzene ring, therefore making it more stable and therefore more reactive compared to electrophilic substitution.
Exercise $7$
The following compound is less reactive towards electrophilic substitution than aniline? Explain.
Answer
As seen in resonance the electron density is also localized off of the ring, thereby deactivating it compared to aniline.
Exercise $8$
Consider the intermediates of the following molecule during an electrophilic substitution. Draw resonance structures for ortho, meta, and para attacks.
Answer | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/16%3A_Electrophilic_Attack_on_Derivatives_of_Benzene%3A_Substituents_Control_Regioselectivity/16.3%3A_Directing__Effects__of_Substituents_in_Conjugation__wi.txt |
Objectives
After completing this section, you should be able to
1. predict the position or positions at which electrophilic substitution will occur when a third substituent is introduced into a disubstituted benzene ring.
2. explain the observed substitution pattern when a third substituent is introduced into a disubstituted benzene ring.
Orientational Interaction of Substituents
When a benzene ring has two substituent groups, each exerts an influence on subsequent substitution reactions. The activation or deactivation of the ring can be predicted more or less by the sum of the individual effects of these substituents. The site at which a new substituent is introduced depends on the orientation of the existing groups and their individual directing effects. We can identify two general behavior categories, as shown in the following table. Thus, the groups may be oriented in such a manner that their directing influences act in concert, reinforcing the outcome; or are opposed (antagonistic) to each other. Note that the orientations in each category change depending on whether the groups have similar or opposite individual directing effects.
Antagonistic or Non-Cooperative Reinforcing or Cooperative
D = Electron Donating Group (ortho/para-directing)
W = Electron Withdrawing Group (meta-directing)
Reinforcing or Cooperative Substitutions
The products from substitution reactions of compounds having a reinforcing orientation of substituents are easier to predict than those having antagonistic substituents. For example, the six equations shown below are all examples of reinforcing or cooperative directing effects operating in the expected manner. Symmetry, as in the first two cases, makes it easy to predict the site at which substitution is likely to occur. Note that if two different sites are favored, substitution will usually occur at the one that is least hindered by ortho groups.
The first three examples have two similar directing groups in a meta-relationship to each other. In examples 4 through 6, oppositely directing groups have an ortho or para-relationship. The major products of electrophilic substitution, as shown, are the sum of the individual group effects. The strongly activating hydroxyl (–OH) and amino (–NH2) substituents favor dihalogenation in examples 5 and 6.
Antagonistic or Non-Cooperative Substitutions
Substitution reactions of compounds having an antagonistic orientation of substituents require a more careful analysis. If the substituents are identical, as in example 1 below, the symmetry of the molecule will again simplify the decision. When one substituent has a pair of non-bonding electrons available for adjacent charge stabilization, it will normally exert the product determining influence, examples 2, 4 & 5, even though it may be overall deactivating (case 2). Case 3 reflects a combination of steric hindrance and the superior innate stabilizing ability of methyl groups relative to other alkyl substituents. Example 6 is interesting in that it demonstrates the conversion of an activating ortho/para-directing group into a deactivating meta-directing "onium" cation [–NH(CH3)2(+) ] in a strong acid environment.
Exercises
Exercise \(1\)
Predict the products of the following reactions:
Answer
16.5: Synthetic Strategies Toward Substituted Benzenes
Objectives
After completing this section, you should be able to
1. design a multistep synthesis which may involve reactions in the alkyl side chain of an alkylbenzene and the electrophilic substitution reactions discussed in this chapter. You should pay particular attention to
1. carrying out the reactions in the correct order.
2. using the most appropriate reagents and conditions.
3. the limitations of certain types of reactions.
2. analyse a proposed multistep synthesis involving aromatic substitution to determine its feasibility, point out any errors in the proposal and identify possible problem areas.
Study Notes
As you can see, designing a multistep synthesis requires an analytical mind and an ability to think logically, as well as a knowledge of organic reactions. The best way to become an expert in designing such syntheses is to get lots of practice by doing plenty of problems.
The ability to plan a successful multi step synthesis of complex molecules is one of the goals of organic chemists. It requires a working knowledge of the uses and limitations of many organic reactions - not only which reactions to use, but when. A few examples follow:
From benzene make m-bromoaniline
In this reaction three reactions are required.
1. A nitration
2. A conversion from the nitro group to an amine
3. A bromination
Because the end product is meta a meta directing group must be utilized. Of the nitro, bromine, and amine group, only the nitro group is meta direction. This means that the first step need to be the nitration and not the bromination. Also, the conversion of the nitro group to an amine must occur last because the amine group is ortho/para direction.
From benzene make p-nitropropylbenzene :
In this reaction three reactions are required.
1. A Friedel Crafts acylation
2. A conversion from the acyl group to an alkane
3. A nitration
Because the propyl group has more than two carbons, it must be added in two steps. A Friedel Crafts acylation followed by a Clemmensen Reduction. Remeber that Friedel Crafts reactions are hindered if the benzene ring is strongly deactivated. This means that the acyl group must go on first. Because the end product is para a para directing group must be utilized. Of the nitro, acyl, and alkane group, only the alkane group is meta direction. This means that the acyl group must be converted to an alkane prior to the nitration step.
Exercises
Exercise \(1\)
How would make the following compounds from benzene?
1. m-bromonitrobenzene
2. m-bromoethylbenzene
Answer
Only one possible synthesis is shown for each compound. There are multiple possibilities.
Exercise \(2\)
There is something wrong with the following reaction, what is it?
Answer
The bromine should be in the meta position. Right now it is in the ortho position, from perhaps having the ethyl group present first and then the having it substituted there. BUT the ethyl group is last to form, and the aldehyde and nitro groups would both encourage a meta substitution.
16.7: Polycyclic Aromatic Hydrocarbons and Cancer
To Your Health: Polycyclic Aromatic Hydrocarbons and Cancer
The intense heating required for distilling coal tar results in the formation of PAHs. For many years, it has been known that workers in coal-tar refineries are susceptible to a type of skin cancer known as tar cancer. Investigations have shown that a number of PAHs are carcinogens. One of the most active carcinogenic compounds, benzopyrene, occurs in coal tar and has also been isolated from cigarette smoke, automobile exhaust gases, and charcoal-broiled steaks. It is estimated that more than 1,000 t of benzopyrene are emitted into the air over the United States each year. Only a few milligrams of benzopyrene per kilogram of body weight are required to induce cancer in experimental animals. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/16%3A_Electrophilic_Attack_on_Derivatives_of_Benzene%3A_Substituents_Control_Regioselectivity/16.4%3A_Electrophilic_Attack__on__Disubstituted_Benzenes.txt |
Objectives
After completing this section, you should be able to
1. provide the IUPAC name of an aldehyde or ketone, given its Kekulé, condensed or shorthand structure.
2. draw the structure of an aldehyde or ketone, given its IUPAC name.
3. draw the structure of the following aldehydes and ketones, given their trivial names: formaldehyde, acetaldehyde, benzaldehyde, acetone, acetophenone, benzophenone.
Study Notes
We only use those trivial names listed under Objective 3, above. We use systematic names in all other cases. For example, the systematic name of the compound shown below is benzenecarbaldehyde, but it has the trivial name of benzaldehyde.
When naming unsaturated aldehydes and ketones, you must give the carbonyl group “priority” over the double bond when you are deciding which end of the carbon chain to begin numbering The carbonyl‑carbon of an aldehyde will always be at the end of the carbon chain in an acyclic compound; and therefore numbering always starts at this carbon. It is for this reason, too, that the number “1” is not required when naming a compound such as 2‑ethyl‑4‑methylpentanal.
The most potent and varied odors are aldehydes. Ketones are widely used as industrial solvents. Aldehydes and ketones contain the carbonyl group. Aldehydes are considered the most important functional group. They are often called the formyl or methanoyl group. Aldehydes derive their name from the dehydration of alcohols. Aldehydes contain the carbonyl group bonded to at least one hydrogen atom. Ketones contain the carbonyl group bonded to two carbon atoms.
Introduction
Aldehydes and ketones are organic compounds which incorporate a carbonyl functional group, C=O. The carbon atom of this group has two remaining bonds that may be occupied by hydrogen, alkyl or aryl substituents. If at least one of these substituents is hydrogen, the compound is an aldehyde. If neither is hydrogen, the compound is a ketone.
Naming Aldehydes
The IUPAC system of nomenclature assigns a characteristic suffix -al to aldehydes. For example, H2C=O is methanal, more commonly called formaldehyde. Since an aldehyde carbonyl group must always lie at the end of a carbon chain, it is always is given the #1 location position in numbering and it is not necessary to include it in the name. There are several simple carbonyl containing compounds which have common names which are retained by IUPAC.
Also, there is a common method for naming aldehydes and ketones. For aldehydes common parent chain names, similar to those used for carboxylic acids, are used and the suffix –aldehyde is added to the end. In common names of aldehydes, carbon atoms near the carbonyl group are often designated by Greek letters. The atom adjacent to the carbonyl function is alpha, the next removed is beta and so on.
If the aldehyde moiety (-CHO) is attached to a ring the suffix –carbaldehyde is added to the name of the ring. The carbon attached to this moiety will get the #1 location number in naming the ring.
Summary of Aldehyde Nomenclature rules
1. Aldehydes take their name from their parent alkane chains. The -e is removed from the end and is replaced with -al.
2. The aldehyde funtional group is given the #1 numbering location and this number is not included in the name.
3. For the common name of aldehydes start with the common parent chain name and add the suffix -aldehyde. Substituent positions are shown with Greek letters.
4. When the -CHO functional group is attached to a ring the suffix -carbaldehyde is added, and the carbon attached to that group is C1.
Example 19.1.1
The IUPAC system names are given on top while the common name is given on the bottom in parentheses.
Naming Ketones
The IUPAC system of nomenclature assigns a characteristic suffix of -one to ketones. A ketone carbonyl function may be located anywhere within a chain or ring, and its position is usually given by a location number. Chain numbering normally starts from the end nearest the carbonyl group. Very simple ketones, such as propanone and phenylethanone do not require a locator number, since there is only one possible site for a ketone carbonyl function
The common names for ketones are formed by naming both alkyl groups attached to the carbonyl then adding the suffix -ketone. The attached alkyl groups are arranged in the name alphabetically.
Summary of Ketone Nomenclature rules
1. Ketones take their name from their parent alkane chains. The ending -e is removed and replaced with -one.
2. The common name for ketones are simply the substituent groups listed alphabetically + ketone.
3. Some common ketones are known by their generic names. Such as the fact that propanone is commonly referred to as acetone.
Example 19.1.2
The IUPAC system names are given on top while the common name is given on the bottom in parentheses.
Naming Aldehydes and Ketones in the Same Molecule
As with many molecules with two or more functional groups, one is given priority while the other is named as a substituent. Because aldehydes have a higher priority than ketones, molecules which contain both functional groups are named as aldehydes and the ketone is named as an "oxo" substituent. It is not necessary to give the aldehyde functional group a location number, however, it is usually necessary to give a location number to the ketone.
Naming Dialdehydes and Diketones
For dialdehydes the location numbers for both carbonyls are omitted because the aldehyde functional groups are expected to occupy the ends of the parent chain. The ending –dial is added to the end of the parent chain name.
Example 19.1.4
For diketones both carbonyls require a location number. The ending -dione or -dial is added to the end of the parent chain.
Naming Cyclic Ketones and Diketones
In cyclic ketones the carbonyl group is assigned location position #1, and this number is not included in the name, unless more than one carbonyl group is present. The rest of the ring is numbered to give substituents the lowest possible location numbers. Remember the prefix cyclo is included before the parent chain name to indicate that it is in a ring. As with other ketones the –e ending is replaced with the –one to indicate the presence of a ketone.
With cycloalkanes which contain two ketones both carbonyls need to be given a location numbers. Also, an –e is not removed from the end but the suffix –dione is added.
Naming Carbonyls and Hydroxyls in the Same Molecule
When and aldehyde or ketone is present in a molecule which also contains an alcohol functional group the carbonyl is given nomenclature priority by the IUPAC system. This means that the carbonyl is given the lowest possible location number and the appropriate nomenclature suffix is included. In the case of alcohols the OH is named as a hydroxyl substituent. However, the l in hydroxyl is generally removed.
Naming Carbonyls and Alkenes in the Same Molecule
When and aldehyde or ketone is present in a molecule which also contains analkene functional group the carbonyl is given nomenclature priority by the IUPAC system. This means that the carbonyl is given the lowest possible location number and the appropriate nomenclature suffix is included.
When carbonyls are included with an alkene the following order is followed:
(Location number of the alkene)-(Prefix name for the longest carbon chain minus the -ane ending)-(an -en ending to indicate the presence of an alkene)-(the location number of the carbonyl if a ketone is present)-(either an –one or and -anal ending).
Remember that the carbonyl has priority so it should get the lowest possible location number. Also, remember that cis/tran or E/Z nomenclature for the alkene needs to be included if necessary.
Aldehydes and Ketones as Fragments
• Alkanoyl is the common name of the fragment, though the older naming, acyl, is still widely used.
• Formyl is the common name of the fragment.
• Acety is the common name of the CH3-C=O- fragment.
Additional Examples of Carbonyl Nomenclature
1) Please give the IUPAC name for each compound:
Answers for Question 1
1. 3,4-dimethylhexanal
2. 5-bromo-2-pentanone
3. 2,4-hexanedione
4. cis-3-pentenal
5. 6-methyl-5-hepten-3-one
6. 3-hydroxy-2,4-pentanedione
7. 1,2-cyclobutanedione
8. 2-methyl-propanedial
9. 3-methyl-5-oxo-hexanal
10. cis-2,3-dihydroxycyclohexanone
11. 3-bromo-2-methylcyclopentanecarboaldehyde
12. 3-bromo-2-methylpropanal
2) Please give the structure corresponding to each name:
A) butanal
B) 2-hydroxycyclopentanone
C) 2,3-pentanedione
D) 1,3-cyclohexanedione
E) 3,4-dihydoxy-2-butanone
F) (E) 3-methyl-2-hepten-4-one
G) 3-oxobutanal
H) cis-3-bromocyclohexanecarboaldehyde
I) butanedial
J) trans-2-methyl-3-hexenal
Answers to question 2: | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/17%3A_Aldehydes_and_Ketones_-_The_Carbonyl_Group/17.01%3A_Naming__the_Aldehydes_and__Ketones.txt |
A carbonyl group is a chemically organic functional group composed of a carbon atom double-bonded to an oxygen atom --> [C=O] The simplest carbonyl groups are aldehydes and ketones usually attached to another carbon compound. These structures can be found in many aromatic compounds contributing to smell and taste.
The Carbonyl Group
C=O is prone to additions and nucleophillic attack because or carbon's positive charge and oxygen's negative charge. The resonance of the carbon partial positive charge allows the negative charge on the nucleophile to attack the Carbonyl group and become a part of the structure and a positive charge (usually a proton hydrogen) attacks the oxygen. Just a reminder, the nucleophile is a good acid therefore "likes protons" so it will attack the side with a positive charge.
Before we consider in detail the reactivity of aldehydes and ketones, we need to look back and remind ourselves of what the bonding picture looks like in a carbonyl. Carbonyl carbons are sp2 hybridized, with the three sp2 orbitals forming soverlaps with orbitals on the oxygen and on the two carbon or hydrogen atoms. These three bonds adopt trigonal planar geometry. The remaining unhybridized 2p orbital on the central carbonyl carbon is perpendicular to this plane, and forms a ‘side-by-side’ pbond with a 2p orbital on the oxygen.
The carbon-oxygen double bond is polar: oxygen is more electronegative than carbon, so electron density is higher on the oxygen side of the bond and lower on the carbon side. Recall that bond polarity can be depicted with a dipole arrow, or by showing the oxygen as holding a partial negative charge and the carbonyl carbon a partial positive charge.
A third way to illustrate the carbon-oxygen dipole is to consider the two main resonance contributors of a carbonyl group: the major form, which is what you typically see drawn in Lewis structures, and a minor but very important contributor in which both electrons in the pbond are localized on the oxygen, giving it a full negative charge. The latter depiction shows the carbon with an empty 2p orbital and a full positive charge.
Some Carbonyl Compounds
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris)
This page explains what aldehydes and ketones are, and looks at the way their bonding affects their reactivity. It also considers their simple physical properties such as solubility and boiling points. Aldehydes and ketones are simple compounds which contain a carbonyl group - a carbon-oxygen double bond. They are simple in the sense that they don't have other reactive groups like -OH or -Cl attached directly to the carbon atom in the carbonyl group - as you might find, for example, in carboxylic acids containing -COOH.
Aldehydes
In aldehydes, the carbonyl group has a hydrogen atom attached to it together with either a second hydrogen atom or, more commonly, a hydrocarbon group which might be an alkyl group or one containing a benzene ring. For the purposes of this section, we shall ignore those containing benzene rings.
Notice that these all have exactly the same end to the molecule. All that differs is the complexity of the other group attached. When you are writing formulae for these, the aldehyde group (the carbonyl group with the hydrogen atom attached) is always written as -CHO - never as COH. That could easily be confused with an alcohol. Ethanal, for example, is written as CH3CHO; methanal as HCHO. The name counts the total number of carbon atoms in the longest chain - including the one in the carbonyl group. If you have side groups attached to the chain, notice that you always count from the carbon atom in the carbonyl group as being number 1.
Ketones
In ketones, the carbonyl group has two hydrocarbon groups attached. Again, these can be either alkyl groups or ones containing benzene rings. Again, we'll concentrated on those containing alkyl groups just to keep things simple. Notice that ketones never have a hydrogen atom attached to the carbonyl group.
Propanone is normally written CH3COCH3. Notice the need for numbering in the longer ketones. In pentanone, the carbonyl group could be in the middle of the chain or next to the end - giving either pentan-3-one or pentan-2-one.
Bonding and reactivity
Oxygen is far more electronegative than carbon and so has a strong tendency to pull electrons in a carbon-oxygen bond towards itself. One of the two pairs of electrons that make up a carbon-oxygen double bond is even more easily pulled towards the oxygen. That makes the carbon-oxygen double bond very highly polar.
The slightly positive carbon atom in the carbonyl group can be attacked by nucleophiles. A nucleophile is a negatively charged ion (for example, a cyanide ion, CN-), or a slightly negatively charged part of a molecule (for example, the lone pair on a nitrogen atom in ammonia, NH3).
During the reaction, the carbon-oxygen double bond gets broken. The net effect of all this is that the carbonyl group undergoes addition reactions, often followed by the loss of a water molecule. This gives a reaction known as addition-elimination or condensation. You will find examples of simple addition reactions and addition-elimination if you explore the aldehydes and ketones menu (link at the bottom of the page). Both aldehydes and ketones contain a carbonyl group. That means that their reactions are very similar in this respect.
Where aldehydes and ketones differ
An aldehyde differs from a ketone by having a hydrogen atom attached to the carbonyl group. This makes the aldehydes very easy to oxidise. For example, ethanal, CH3CHO, is very easily oxidised to either ethanoic acid, CH3COOH, or ethanoate ions, CH3COO-.
Ketones don't have that hydrogen atom and are resistant to oxidation. They are only oxidised by powerful oxidising agents which have the ability to break carbon-carbon bonds. You will find the oxidation of aldehydes and ketones discussed if you follow a link from the aldehydes and ketones menu (see the bottom of this page).
Boiling Points
Methanal is a gas (boiling point -21°C), and ethanal has a boiling point of +21°C. That means that ethanal boils at close to room temperature. The other aldehydes and the ketones are liquids, with boiling points rising as the molecules get bigger. The size of the boiling point is governed by the strengths of the intermolecular forces.
• Van der Waals dispersion forces: These attractions get stronger as the molecules get longer and have more electrons. That increases the sizes of the temporary dipoles that are set up. This is why the boiling points increase as the number of carbon atoms in the chains increases - irrespective of whether you are talking about aldehydes or ketones.
• van der Waals dipole-dipole attractions: Both aldehydes and ketones are polar molecules because of the presence of the carbon-oxygen double bond. As well as the dispersion forces, there will also be attractions between the permanent dipoles on nearby molecules. That means that the boiling points will be higher than those of similarly sized hydrocarbons - which only have dispersion forces. It is interesting to compare three similarly sized molecules. They have similar lengths, and similar (although not identical) numbers of electrons.
molecule type boiling point (°C)
CH3CH2CH3 alkane -42
CH3CHO aldehyde +21
CH3CH2OH alcohol +78
Notice that the aldehyde (with dipole-dipole attractions as well as dispersion forces) has a boiling point higher than the similarly sized alkane which only has dispersion forces. However, the aldehyde's boiling point isn't as high as the alcohol's. In the alcohol, there is hydrogen bonding as well as the other two kinds of intermolecular attraction.
Although the aldehydes and ketones are highly polar molecules, they don't have any hydrogen atoms attached directly to the oxygen, and so they can't hydrogen bond with each other.
Solubility in water
The small aldehydes and ketones are freely soluble in water but solubility falls with chain length. For example, methanal, ethanal and propanone - the common small aldehydes and ketones - are miscible with water in all proportions.The reason for the solubility is that although aldehydes and ketones can't hydrogen bond with themselves, they can hydrogen bond with water molecules. One of the slightly positive hydrogen atoms in a water molecule can be sufficiently attracted to one of the lone pairs on the oxygen atom of an aldehyde or ketone for a hydrogen bond to be formed.
There will also, of course, be dispersion forces and dipole-dipole attractions between the aldehyde or ketone and the water molecules. Forming these attractions releases energy which helps to supply the energy needed to separate the water molecules and aldehyde or ketone molecules from each other before they can mix together.
As chain lengths increase, the hydrocarbon "tails" of the molecules (all the hydrocarbon bits apart from the carbonyl group) start to get in the way. By forcing themselves between water molecules, they break the relatively strong hydrogen bonds between water molecules without replacing them by anything as good. This makes the process energetically less profitable, and so solubility decreases. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/17%3A_Aldehydes_and_Ketones_-_The_Carbonyl_Group/17.02%3A_Structure_of_the_Carbonyl__Group.txt |
Objectives
After completing this section, you should be able to
1. identify the region of the infrared spectrum in which the carbonyl absorption of aldehydes and ketones is found.
2. identify the region of the infrared spectrum in which the two characteristic C$\ce{-}$H absorptions of aldehydes are found.
3. use a table of characteristic absorption frequencies to assist in the determination of the structure of an unknown aldehyde or ketone, given its infrared spectrum and other spectral or experimental data.
4. identify the region of a proton NMR spectrum in which absorptions caused by the presence of aldehydic protons and protons attached to the α‑carbon atoms of aldehydes and ketones occur.
5. identify two important fragmentations that occur when aliphatic aldehydes and ketones are subjected to analysis by mass spectrometry.
Key Terms
Make certain that you can define, and use in context, the key term below.
• McLafferty rearrangement
Study Notes
The appearance of a strong absorption at 1660–1770 cm−1 in the infrared spectrum of a compound is a clear indication of the presence of a carbonyl group. Although you need not remember the detailed absorptions it is important that you realize that the precise wavenumber of the infrared absorption can often provide some quite specific information about the environment of the carbonyl group in a compound. Notice how conjugation between a carbonyl group and a double bond (α, β‑unsaturated aldehyde or ketone or aromatic ring) lowers the absorption by about 25–30 cm−1.
You may wish to review the McLafferty rearrangement and the alpha cleavage in Section 12.3.
IR Spectra
The carbonyl stretching vibration band C=O of saturated aliphatic ketones appears:
C=O stretch
• aliphatic ketones 1715 cm-1
• alpha, beta-unsaturated ketones 1685-1666 cm-1
Figure 8. shows the spectrum of 2-butanone. This is a saturated ketone, and the C=O band appears at 1715.
Figure 8. Infrared Spectrum of 2-Butanone
If a compound is suspected to be an aldehyde, a peak always appears around 2720 cm-1 which often appears as a shoulder-type peak just to the right of the alkyl C–H stretches.
H–C=O stretch 2830-2695 cm-1
C=O stretch
• aliphatic aldehydes 1740-1720 cm-1
• alpha, beta-unsaturated aldehydes 1710-1685 cm-1
Figure 9. shows the spectrum of butyraldehyde.
Figure 9. Infrared Spectrum of Butyraldehyde
NMR Spectra
Hydrogens attached to carbon adjacent to the sp2 hybridized carbon in aldehydes and ketones usually show up 2.0-2.5 ppm.
.
Aldehyde hydrogens are highly deshielded and appear far downfield as 9-10 ppm.
Chemical shift of each protons is predicted by 1H chemical shift ranges (Ha): chemical shift of methyl groups (1.1 ppm). (Hb) The chemical shift of the -CH- group move downfield due to effect an adjacent aldehyde group: (2.4 ppm). The chemical shift of aldehyde hydrogen is highly deshielded (9.6 ppm).
4) Splitting pattern is determined by (N+1) rule: Ha is split into two peaks by Hb(#of proton=1). Hb has the septet pattern by Ha (#of proton=6). Hc has one peak.(Note that Hc has doublet pattern by Hb due to vicinal proton-proton coupling.)
17.04: Preparation of Aldehydes and Ketones
Objectives
After completing this section, you should be able to
1. describe in detail the methods for preparing aldehydes discussed in earlier units (i.e., the oxidation of primary alcohols and the cleavage of alkenes).
2. write an equation to describe the reduction of an ester to an aldehyde.
1. identify the product formed when a given ester is reduced with diisobutylaluminum hydride.
2. identify the reagents and conditions used in the reduction of an ester to an aldehyde.
3. identify the disadvantages of using diisobutylaluminum hydride to reduce an ester to an aldehyde.
3. describe in detail the methods for preparing ketones discussed in earlier units (i.e., the oxidation of secondary alcohols, the ozonolysis of alkenes, Friedel‑Crafts acylation, and the hydration of terminal alkynes).
1. write an equation to illustrate the formation of a ketone through the reaction of an acid chloride with a dialkylcopper lithium reagent.
2. identify the ketone produced from the reaction of a given acid chloride with a specified dialkylcopper lithium reagent.
3. identify the acid chloride, the dialkylcopper lithium reagent, or both, needed to prepare a specific ketone.
Study Notes
You may wish to review the sections in which we discuss the oxidation of alcohols (17.7) and the cleavage of alkenes (8.8). A third method of preparing aldehydes is to reduce a carboxylic acid derivative; for example, to reduce an ester with diisobutylaluminum hydride (DIBAL‑H).
There are essentially five methods of preparing ketones in the laboratory. Four of them have been discussed in earlier sections:
1. the oxidation of a secondary alcohol—Section 17.7.
2. the ozonolysis of an alkene—Section 8.8.
3. Friedel‑Crafts acylation—Section 16.3.
4. the hydration of a terminal alkyne—Section 9.4.
The “new” method we introduce in this section involves the reaction of an acid chloride with a diorganocopper reagent. The latter substances were discussed in Section 10.7, which you might now wish to review.
Aldehydes and ketones can be prepared using a wide variety of reactions. Although these reactions are discussed in greater detail in other sections, they are listed here as a summary and to help with planning multistep synthetic pathways. Please use the appropriate links to see more details about the reactions.
Hydration of an alkyne to form aldehydes
Anti-Markovnikov addition of a hydroxyl group to an alkyne forms an aldehyde. The addition of a hydroxyl group to an alkyne causes tautomerization which subsequently forms a carbonyl.
Oxidation of 2o alcohols to form ketones
Typically uses Jones reagent (CrO3 in H2SO4) but many other reagents can be used
Hydration of an alkyne to form ketones
The addition of a hydroxyl group to an alkyne causes tautomerization which subsequently forms a carbonyl. Markovnikov addition of a hydroxyl group to an alkyne forms a ketone.
Alkenes can be cleaved using ozone (O3) to form aldehydes and/or ketones
This is an example of a Ozonolysis reaction.
Example 1: | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/17%3A_Aldehydes_and_Ketones_-_The_Carbonyl_Group/17.03%3A_Spectroscopic_Properties_of_Aldehydes_and__Ketones.txt |
Nucleophilic Addition to Aldehydes and Ketones
The result of carbonyl bond polarization, however it is depicted, is straightforward to predict. The carbon, because it is electron-poor, is an electrophile: it is a great target for attack by an electron-rich nucleophilic group. Because the oxygen end of the carbonyl double bond bears a partial negative charge, anything that can help to stabilize this charge by accepting some of the electron density will increase the bond’s polarity and make the carbon more electrophilic. Very often a general acid group serves this purpose, donating a proton to the carbonyl oxygen.
The same effect can also be achieved if a Lewis acid, such as a magnesium ion, is located near the carbonyl oxygen.
Unlike the situation in a nucleophilic substitution reaction, when a nucleophile attacks an aldehyde or ketone carbon there is no leaving group – the incoming nucleophile simply ‘pushes’ the electrons in the pi bond up to the oxygen.
Alternatively, if you start with the minor resonance contributor, you can picture this as an attack by a nucleophile on a carbocation.
After the carbonyl is attacked by the nucleophile, the negatively charged oxygen has the capacity to act as a nucleophile. However, most commonly the oxygen acts instead as a base, abstracting a proton from a nearby acid group in the solvent or enzyme active site.
This very common type of reaction is called a nucleophilic addition. In many biologically relevant examples of nucleophilic addition to carbonyls, the nucleophile is an alcohol oxygen or an amine nitrogen, or occasionally a thiol sulfur. In one very important reaction type known as an aldol reaction (which we will learn about in section 13.3) the nucleophile attacking the carbonyl is a resonance-stabilized carbanion. In this chapter, we will concentrate on reactions where the nucleophile is an oxygen or nitrogen.
Nucleophilic Substitution of RCOZ (Z = Leaving Group)
Carbonyl compounds with leaving groups have reactions similar to aldehydes and ketones. The main difference is the presence of an electronegative substituent that can act as a leaving group during a nucleophile substitution reaction. Although there are many types of carboxylic acid derivatives known, this article focuses on four: acid halides, acid anhydrides, esters, and amides.
General mechanism
1) Nucleophilic attack on the carbonyl
2) Leaving group is removed
Although aldehydes and ketones also contain carbonyls, their chemistry is distinctly different because they do not contain suitable leaving groups. Once a tetrahedral intermediate is formed, aldehydes and ketones cannot reform their carbonyls. Because of this, aldehydes and ketones typically undergo nucleophilic additions and not substitutions.
The relative reactivity of carboxylic acid derivatives toward nucleophile substitutions is related to the electronegative leaving group’s ability to activate the carbonyl. The more electronegative leaving groups withdraw electron density from the carbonyl, thereby increasing its electrophilicity.
17.06: Addition of Water to Form Hydrates
Contributors and Attributions
Prof. Steven Farmer (Sonoma State University)
William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry
17.07: Addition of Alcohols to Form Hemiacetals and Acetals
Objectives
After completing this section, you should be able to
1. write an equation to illustrate the formation of acetals.
2. identify the acetal formed from the reaction of a given aldehyde or ketone with a given alcohol.
3. identify the carbonyl compound, the alcohol, or both, needed to form a given acetal.
4. write a detailed mechanism for the reaction which occurs between an aldehyde or a ketone and an alcohol.
5. explain how an acid catalyst makes aldehydes and ketones more susceptible to attack by alcohols.
6. illustrate how the reversibility of the reaction between an aldehyde or a ketone and an alcohol can be used to protect a carbonyl group during an organic synthesis.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• acetal
• hemiacetal
Study Notes
This section presents a second example of the use of a protecting group. [The first was in the discussion of alcohols, Section 17.8.] Because of the reactivity of hydroxy groups and carbonyl groups, we often need to protect such groups during organic syntheses. When you are designing multi‑step syntheses as part of an assignment or examination question, you must always keep in mind the possibility that you may need to protect such groups to carry out the desired sequence of reactions successfully.
In this organic chemistry topic, we shall see how alcohols (R-OH) add to carbonyl groups. Carbonyl groups are characterized by a carbon-oxygen double bond. The two main functional groups that consist of this carbon-oxygen double bond are Aldehydes and Ketones.
Introduction
It has been demonstrated that water adds rapidly to the carbonyl function of aldehydes and ketones to form geminal-diol. In a similar reaction alcohols add reversibly to aldehydes and ketones to form hemiacetals (hemi, Greek, half). This reaction can continue by adding another alcohol to form an acetal. Hemiacetals and acetals are important functional groups because they appear in sugars.
To achieve effective hemiacetal or acetal formation, two additional features must be implemented. First, an acid catalyst must be used because alcohol is a weak nucleophile; and second, the water produced with the acetal must be removed from the reaction by a process such as a molecular sieves or a Dean-Stark trap. The latter is important, since acetal formation is reversible. Indeed, once pure hemiacetal or acetals are obtained they may be hydrolyzed back to their starting components by treatment with aqueous acid and an excess of water.
Formation of Acetals
Acetals are geminal-diether derivatives of aldehydes or ketones, formed by reaction with two equivalents (or an excess amount) of an alcohol and elimination of water. Ketone derivatives of this kind were once called ketals, but modern usage has dropped that term. It is important to note that a hemiacetal is formed as an intermediate during the formation of an acetal.
Mechanism for Hemiacetal and Acetal Formation
The mechanism shown here applies to both acetal and hemiacetal formation
1) Protonation of the carbonyl
2) Nucleophilic attack by the alcohol
3) Deprotonation to form a hemiacetal
4) Protonation of the alcohol
5) Removal of water
6) Nucleophilic attack by the alcohol
7) Deprotonation by water
Acetals as Protecting Groups
The importance of acetals as carbonyl derivatives lies chiefly in their stability and lack of reactivity in neutral to strongly basic environments. As long as they are not treated by acids, especially aqueous acid, acetals exhibit all the lack of reactivity associated with ethers in general. Among the most useful and characteristic reactions of aldehydes and ketones is their reactivity toward strongly nucleophilic (and basic) metallo-hydride, alkyl and aryl reagents. If the carbonyl functional group is converted to an acetal these powerful reagents have no effect; thus, acetals are excellent protective groups, when these irreversible addition reactions must be prevented.
In the following example we would like a Grignard reagent to react with the ester and not the ketone. This cannot be done without a protecting group because Grignard reagents react with esters and ketones. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/17%3A_Aldehydes_and_Ketones_-_The_Carbonyl_Group/17.05%3A_Reactivity_of_the_Carbonyl__Group%3A_Mechanisms_of_Addition.txt |
Acetals as Protecting Groups
The importance of acetals as carbonyl derivatives lies chiefly in their stability and lack of reactivity in neutral to strongly basic environments. As long as they are not treated by acids, especially aqueous acid, acetals exhibit all the lack of reactivity associated with ethers in general. Among the most useful and characteristic reactions of aldehydes and ketones is their reactivity toward strongly nucleophilic (and basic) metallo-hydride, alkyl and aryl reagents. If the carbonyl functional group is converted to an acetal these powerful reagents have no effect; thus, acetals are excellent protective groups, when these irreversible addition reactions must be prevented.
In the following example we would like a Grignard reagent to react with the ester and not the ketone. This cannot be done without a protecting group because Grignard reagents react with esters and ketones.
Contributors
Compound 1.4.1.4A illustrates several important points in Protection / Deprotection protocol. Both the functional groups could react with a Grignard Reagent. Carboxylic acid group would first react with one mole of the Grignard Reagent to give a carboxylate anion salt. This anion does not react any further with the reagent. When two moles of Grignard Reagent are added to the reaction mixture, the second mole attacks the ketone to give a tertiary alcohol. On aqueous work-up, the acid group is regenerated. Thus, the first mole of the reagent provides a selective transient protection for the –COOH group. Once the acid group is esterified, such selectivity towards this reagent is lost. The reagent attacks at both sites. If reaction is desired only at the ester site, the keto- group should be selectively protected as an acetal. In the next step, the grignard reaction is carried out. Now the reagent has only one group available for reaction. On treatment with acid, the ketal protection in the intermediate compound is also hydrolyzed to regenerated the keto- group.
Fig 1.4.1.4
Protection of Aldehydes and Ketones
Since alcohols, aldehydes and ketones are the most frequently manipulated functional groups in organic synthesis, a great deal of work has appeared in their protection / deprotection strategies. In this discussion let us focus on the classes of protecting groups rather than an exhaustive treatment of all the protections.
Acetals
There are two general methods for the introduction of this protection. Transketalation is the method of choice when acetals (ketals) with methanol are desired. Acetone is the by-product, which has to be removed to shift the equilibrium to the right hand side. This is achieved by refluxing with a large excess of the acetonide reagent. Acetone formed is constantly distilled. In the case of cyclic diols, the water formed is continuously removed using a Dean-Stork condenser (Fig 1.4.1.6).
Fig 1.4.1.6
The rate of formation of ketals from ketones and 1,2-ethanediol (ethylene glycol), 1,3-propanediol and 2,2-dimethyl-1,3-propanediol are different. So is the deketalation reaction. This has enabled chemists to selectively work at one center. The following examples from steroid chemistry illustrate these points (Fig 1.4.1.7).
Fig 1.4.1.7
The demand for Green Chemistry processes has prompted search for new green procedures. Some examples from recent literature are given here (Fig 1.4.1.8).
Fig 1.4.1.8
Thioketals
Compared with their oxygen analogues, thioketals markedly differ in their chemistry. The formation as well as deprotection is promoted by suitable Lewis acids. The thioacetals are markedly stable under deketalation conditions, thus paving way for selective operations at two different centers. When conjugated ketones are involved, the ketal formation (as well as deprotection) proceeds with double bond migration. On the other hand, thioketals are formed and deketalated without double bond migration (Fig 1.4.1.9).
Fig 1.4.1.9
Silyl Ethers (R – OSiR3)
The oxygen – silicon sigma bond is stable to lithium and Grignard reagents, nucleophiles and hydride reagents but very unstable to water and mild aqueous acid and base conditions. A silyl ether of secondary alcohol is less reactive than that of a primary alcohol. The O – trimethylsilyl (O – SiMe3) was first protection of this class. (Fig 1.4.1.24).
Fig 1.4.1.24
Replacement of methyl group with other alkyl and aryl groups gives a large variety of silyl ether with varying degrees of stability towards hydrolysis (Fig 1.4.1.25).
Fig 1.4.1.25
The following examples illustrate the selectivity in formation and hydrolysis of this group (Fig 1.4.1.26).
Fig 1.4.1.26
Contributors
• Prof. R Balaji Rao (Department of Chemistry, Banaras Hindu University, Varanasi) as part of Information and Communication Technology | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/17%3A_Aldehydes_and_Ketones_-_The_Carbonyl_Group/17.08%3A_Acetals__as_Protecting_Groups.txt |
Objectives
After completing this section, you should be able to
1. write equations to describe the reactions that occur between aldehydes or ketones and primary or secondary amines.
2. identify the product formed from the reaction of a given aldehyde or ketone with a given primary or secondary amine.
3. identify the aldehyde or ketone, the amine, or both, required in the synthesis of a given imine or enamine.
4. write the detailed mechanism for the reaction of an aldehyde or ketone with a primary amine.
5. write the detailed mechanism for the reaction of an aldehyde or ketone with a secondary amine.
6. explain why the rate of a reaction between an aldehyde or ketone and a primary or secondary amine is dependent on pH.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• 2,4‑dinitrophenylhydrozone
• enamine
• imine
Study Notes
An imine is a compound that contains the structural unit
An enamine is a compound that contains the structural unit
Both of these types of compound can be prepared through the reaction of an aldehyde or ketone with an amine.
You may have the opportunity to observe the reaction of an aldehyde and ketone with 2,4‑dinitrophenylhydrazine (Brady’s reagent) to form a 2,4‑dinitrophenylhydrozone in the laboratory. This is a classical organic chemistry test to confirm the presence of a carbonyl group. The reaction produces very colourful and bright precipitates of yellow, orange and red.
If you can understand why the two reactions of imine and enamine formation are essentially identical, and can write a detailed mechanism for each one, you are well on the way to mastering organic chemistry. If you understand how and why these reactions occur, you can keep the amount of material that you need to memorize to a minimum.
Reaction with Primary Amines to form Imines
The reaction of aldehydes and ketones with ammonia or 1º-amines forms imine derivatives, also known as Schiff bases (compounds having a C=N function). Water is eliminated in the reaction, which is acid-catalyzed and reversible in the same sense as acetal formation. The pH for reactions which form imine compounds must be carefully controlled. The rate at which these imine compounds are formed is generally greatest near a pH of 5, and drops at higher and lower pH's. At high pH there will not be enough acid to protonate the OH in the intermediate to allow for removal as H2O. At low pH most of the amine reactant will be tied up as its ammonium conjugate acid and will become non-nucleophilic.
Converting reactants to products simply
Mechanism of imine formation
1) Nucleophilic attack
2) Proton transfer
3) Protonation of OH
4) Removal of water
5) Deprotonation
Reversibility of imine forming reactions
Imines can be hydrolyzed back to the corresponding primary amine under acidic conditons.
Reactions involving other reagents of the type Y-NH2
Imines are sometimes difficult to isolate and purify due to their sensitivity to hydrolysis. Consequently, other reagents of the type Y–NH2 have been studied, and found to give stable products (R2C=N–Y) useful in characterizing the aldehydes and ketones from which they are prepared. Some of these reagents are listed in the following table, together with the structures and names of their carbonyl reaction products. Hydrazones are used as part of the Wolff-Kishner reduction and will be discussed in more detail in another module.
With the exception of unsubstituted hydrazones, these derivatives are easily prepared and are often crystalline solids - even when the parent aldehyde or ketone is a liquid. Since melting points can be determined more quickly and precisely than boiling points, derivatives such as these are useful for comparison and identification of carbonyl compounds. It should be noted that although semicarbazide has two amino groups (–NH2) only one of them is a reactive amine. The other is amide-like and is deactivated by the adjacent carbonyl group.
Problems
1)Please draw the products of the following reactions.
2) Please draw the structure of the reactant needed to produce the indicated product.
3) Please draw the products of the following reactions.
1)
2)
3)
Contributors
Prof. Steven Farmer (Sonoma State University)
William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry
Reaction with Secondary Amines to form Enamines
Most aldehydes and ketones react with 2º-amines to give products known as enamines. It should be noted that, like acetal formation, these are acid-catalyzed reversible reactions in which water is lost. Consequently, enamines are easily converted back to their carbonyl precursors by acid-catalyzed hydrolysis.
Mechanism
1) Nuleophilic attack
2) Proton transfer
3) Protonation of OH
4) Removal of water
5) Deprotonation
Problems
1) Please draw the products for the following reactions.
2) Please give the structure of the reactant needed to product the following product
1)
2) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/17%3A_Aldehydes_and_Ketones_-_The_Carbonyl_Group/17.09%3A_Nucleophilic_Addition_of_Ammonia_and__Its__Derivatives.txt |
Objectives
After completing this section, you should be able to
1. write an equation to illustrate the Wolff‑Kishner reduction of an aldehyde or ketone.
2. identify the product formed from the Wolff‑Kishner reduction of a given aldehyde or ketone.
Key Terms
Make certain that you can define, and use in context, the key term below.
• Wolff‑Kishner reduction
Study Notes
After studying this section, you can add yet another method of reducing organic compounds to your growing list of reduction reactions.
Aldehydes and ketones can be converted to a hydrazine derivative by reaction with hydrazine. These "hydrazones" can be further converted to the corresponding alkane by reaction with base and heat. These two steps can be combined into one reaction called the Wolff-Kishner Reduction which represents a general method for converting aldehydes and ketones into alkanes. Typically a high boiling point solvent, such as ethylene glycol, is used to provide the high temperatures needed for this reaction to occur. Note! Nitrogen gas is produced as part of this reaction.
Example
Mechanism of the Wolff-Kishner Reduction
1) Deprotonation of Nitrogen
2) Protonation of the Carbon
3) Deprotonation of Nitrogen
4) Protonation of Carbon
Problems
1) Please draw the products of the following reactions.
1)
17.11: Addition of Hydrogen Cyanide to Give Cyanohydrins
Objectives
After completing this section, you should be able to
1. write an equation to describe the formation of a cyanohydrin from an aldehyde or ketone.
2. identify the cyanohydrin formed from the reaction of a given aldehyde or ketone with hydrogen cyanide.
3. identify the aldehyde or ketone, the reagents, or both, needed to prepare a given cyanohydrin.
4. write the detailed mechanism for the addition of hydrogen cyanide to an aldehyde or ketone.
Key Terms
Make certain that you can define, and use in context, the key term below.
• cyanohydrin
Study Notes
For successful cyanohydrin formation it is important to have free cyanide ions available to react with the ketone or aldehyde. This can be achieved by using a salt (e.g. KCN or NaCN) or a silylated (e.g. Me3SiCN) form of cyanide under acidic conditions or by using HCN with some base added to produce the needed CN nucleophile.
Cyanohydrins have the structural formula of R2C(OH)CN. The “R” on the formula represents an alkyl, aryl, or hydrogen. To form a cyanohydrin, a hydrogen cyanide adds reversibly to the carbonyl group of an organic compound thus forming a hydroxyalkanenitrile adducts (commonly known and called as cyanohydrins).
Figure 19.6.1: General structure of a cyanohydrin
Hydrogen cyanide adds across the carbon-oxygen double bond in aldehydes and ketones to produce compounds known as hydroxynitriles. For example, with ethanal (an aldehyde) you get 2-hydroxypropanenitrile:
With propanone (a ketone) you get 2-hydroxy-2-methylpropanenitrile:
The reaction isn't normally done using hydrogen cyanide itself, because this is an extremely poisonous gas. Instead, the aldehyde or ketone is mixed with a solution of sodium or potassium cyanide in water to which a little sulphuric acid has been added. The pH of the solution is adjusted to about 4 - 5, because this gives the fastest reaction. The solution will contain hydrogen cyanide (from the reaction between the sodium or potassium cyanide and the sulphuric acid), but still contains some free cyanide ions. This is important for the mechanism.
Mechanism of Cyanohydrin Formation
Acid-catalyzed hydrolysis of silylated cyanohydrins has recently been shown to give cyanohydrins instead of ketones; thus an efficient synthesis of cyanohydrins has been found which works with even highly hindered ketones.
Acetone Cyanohydrins
Acetone cyanohydrins (ACH) have the structural formula of (CH3)2C(OH)CN. It is an organic compound serves in the production of methyl methacrylate (also known as acrylic).
Figure 19.6.2: Acetone cyanohydrins
It is classified as an extremely hazardous substance, since it rapidly decomposes when it's in contact with water. In ACH, sulfuric acid is treated to give the sulfate ester of the methacrylamid. Preparations of other cyanohydrins are also used from ACH: for HACN to Michael acceptors and for the formylation of arenas. The treatment with lithium hydride affords anhydrous lithium cyanide.
Figure 19.6.3: Reduction of Acetone cyanohydrins
Other Cyanohydrins
Other cyanohydrins, excluding acetone cyanohydrins, are: mandelonitrile and glycolonitrile.
Figure 19.6.4: Structures of Madelonitrile (left) and glycolonitrile (right)
Mandelonitrile have a structural formula of C6H5CH(OH)CN and occur in pits of some fruits. Glycolonitrile is an organic compound with the structural formula of HOCH2CN, which is the simplest cyanohydrin that is derived by formaldehydes. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/17%3A_Aldehydes_and_Ketones_-_The_Carbonyl_Group/17.10%3A_Deoxygenation_of_the_Carbonyl_Group.txt |
Objectives
After completing this section, you should be able to
1. write an equation to illustrate the formation of an ylide (phosphorane).
2. write an equation to illustrate the reaction that takes place between an ylide and an aldehyde or ketone.
3. identify the alkene which results from the reaction of a given ylide with a given aldehyde or ketone.
4. identify the aldehyde or ketone, the ylide, or both, needed to prepare a given alkene by a Wittig reaction.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• betaine
• Wittig reaction
• ylide (phosphorane)
Study Notes
The name triphenylphosphine is derived as follows: the compound PH3 is called phosphine; replacing the three hydrogen atoms with phenyl groups therefore gives us triphenylphosphine.
Note the following series of IUPAC‑accepted trivial names:
NH3—ammonia
PH3—phosphine
AsH3—arsine
Organophosphorus ylides react with aldehydes or ketones to give substituted alkenes in a transformation called the Wittig reaction. This reaction is named for George Wittig who was awarded the Nobel prize for this work in 1979. A principal advantage of alkene synthesis by the Wittig reaction is that the location of the double bond is absolutely fixed, in contrast to the mixtures often produced by alcohol dehydration.
Preparation of Phosphorus Ylides
It has been noted that dipolar phosphorus compounds are stabilized by p-d bonding. This bonding stabilization extends to carbanions adjacent to phosphonium centers, and the zwitterionic conjugate bases derived from such cations are known as ylides. An ylide is defined as a compound with opposite charges on adjacent atoms both of which have complete octets. For the Wittig reaction discussed below an organophosphorus ylide, also called Wittig reagents, will be used. The ability of phosphorus to hold more than eight valence electrons allows for a resonance structure to be drawn forming a double bonded structure.
The stabilization of the carbanion provided by the phosphorus causes an increase in acidity (pKa ~35). Very strong bases, such as butyl lithium, are required for complete formation of ylides.
The ylides shown here are all strong bases. Like other strongly basic organic reagents, they are protonated by water and alcohols, and are sensitive to oxygen. Water decomposes phosphorous ylides to hydrocarbons and phosphine oxides, as shown.
Although many ylides are commercially available it is often necessary to create them synthetically. Ylides can be synthesized from an alkyl halide and a trialkyl phosphine. Typically triphenyl phosphine is used to synthesize ylides. Because a SN2 reaction is used in the ylide synthesis methyl and primary halides perform the best. Secondary halides can also be used but the yields are generally lower. This should be considered when planning out a synthesis which involves a synthesized Wittig reagent.
1) SN2 reaction
2) Deprotonation
The Wittig Reaction
The most important use of ylides in synthesis comes from their reactions with aldehydes and ketones, which are initiated in every case by a covalent bonding of the nucleophilic alpha-carbon to the electrophilic carbonyl carbon. Ylides react to give substituted alkenes in a transformation called the Wittig reaction. This reaction is named for George Wittig who was awarded the Nobel prize for this work in 1979. A principal advantage of alkene synthesis by the Wittig reaction is that the location of the double bond is absolutely fixed, in contrast to the mixtures often produced by alcohol dehydration.
Going from reactants to products simplified
Mechanism of the Wittig reaction
Following the initial carbon-carbon bond formation, two intermediates have been identified for the Wittig reaction, a dipolar charge-separated species called a betaine and a four-membered heterocyclic structure referred to as an oxaphosphatane. Cleavage of the oxaphosphatane to alkene and phosphine oxide products is exothermic and irreversible.
1) Nucleophillic attack on the carbonyl
2) Formation of a 4 membered ring
3) Formation of the alkene
Limitation of the Wittig reaction
If possible both E and Z isomer of the double bond will be formed. This should be considered when planning a synthesis involving a Wittig Reaction.
Problems
1) Please write the product of the following reactions.
2) Please indicate the starting material required to produce the product.
3) Please draw the structure of the oxaphosphetane which is made during the mechanism of the reaction given that produces product C.
4) Please draw the structure of the betaine which is made during the mechanism of the reaction given that produces product D.
5) Please give a detailed mechanism and the final product of this reaction
6) It has been shown that reacting and epoxide with triphenylphosphine forms an alkene. Please propose a mechanism for this reaction. Review the section on epoxide reactions if you need help.
Answers
1)
2)
3)
4)
5)
Nucleophilic attack on the carbonyl
Formation of a 4 membered ring
Formation of the alkene
6) Nucleophillic attack on the epoxide
Formation of a 4 membered ring
Formation of the alkene
17.13: Oxidation by Peroxycarboxylic Acids: The Baeyer- Villiger Oxidation
Baeyer-Villiger oxidation is the oxidation of a ketone to a carboxylic acid ester using a peroxyacid as the oxidizing agent. eg. 1:
eg. 2:
Mechanism
When the two ligands on the carbonyl carbon in the ketone are different, Baeyer-Villiger oxidation is regioselective. Of the two alpha carbons in the ketone, the one that can stabilize a positive charge more effectively, which is the more highly substituted one, migrates from carbon to oxygen preferentially.
eg. 1:
eg. 2:
17.14: Oxidative Chemical Tests for Aldehydes
Objectives
After completing this section, you should be able to
1. write an equation for the oxidation of an aldehyde using
1. CrO3/sulphuric acid.
2. Tollens reagent.
2. explain the difference in structure which makes aldehydes susceptible to oxidation and ketones difficult to oxidize.
3. identify the carboxylic acid produced when a given aldehyde is oxidized.
4. identify the aldehyde, the oxidizing agent, or both, needed to prepare a given carboxylic acid.
Key Terms
Make certain that you can define, and use in context, the key term below.
• Tollens reagent
Study Notes
An important difference between aldehydes and ketones is the ease with which the latter can be oxidized. Tollen’s reagent is a classical organic laboratory technique to test for the presence of an aldehyde. The reagent consists of silver(I) ions dissolved in dilute ammonia. When the aldehyde is oxidized, the silver(I) ions are reduced to silver metal. When the reaction is carried out in a test‑tube, the metallic silver is deposited on the walls of the tube, giving it a mirrorlike appearance. This characteristic accounts for the term “silver mirror test” which is applied when this reaction is used to distinguish between aldehydes and ketones—the latter, of course, do not react.
This page looks at ways of distinguishing between aldehydes and ketones using oxidizing agents such as acidified potassium dichromate(VI) solution, Tollens' reagent, Fehling's solution and Benedict's solution.
Why do aldehydes and ketones behave differently?
You will remember that the difference between an aldehyde and a ketone is the presence of a hydrogen atom attached to the carbon-oxygen double bond in the aldehyde. Ketones don't have that hydrogen.
Oxidation of Aldehydes
The presence of that hydrogen atom makes aldehydes very easy to oxidize. Or, put another way, they are strong reducing agents. The most common reagent for this conversion is CrO3 in aqueous acid. This reaction generally gives good yields at room temperature.
Unfortunately, the acid condition for the previous reaction can cause unwanted side reaction. If this problem occurs it can be rectified by using a solution of sliver oxide, Ag2O, in aqueous ammonia, also called Tollens' reagent.
Because ketones do not have that particular hydrogen atom, they are resistant to oxidation, and only very strong oxidizing agents like potassium manganate(VII) solution (potassium permanganate solution) oxidize ketones. However, they do it in a destructive way, breaking carbon-carbon bonds and forming two carboxylic acids. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/17%3A_Aldehydes_and_Ketones_-_The_Carbonyl_Group/17.12%3A_Addition__of__Phosphorus_Ylides%3A__The__Wittig__Reaction.txt |
For alkylation reactions of enolate anions to be useful, these intermediates must be generated in high concentration in the absence of other strong nucleophiles and bases. The aqueous base conditions used for the aldol condensation are not suitable because the enolate anions of simple carbonyl compounds are formed in very low concentration, and hydroxide or alkoxide bases induce competing SN2 and E2 reactions of alkyl halides. It is necessary, therefore, to achieve complete conversion of aldehyde or ketone reactants to their enolate conjugate bases by treatment with a very strong base (pKa > 25) in a non-hydroxylic solvent before any alkyl halides are added to the reaction system. Some bases that have been used for enolate anion formation are: NaH (sodium hydride, pKa > 45), NaNH2 (sodium amide, pKa = 34), and LiN[CH(CH3)2]2 (lithium diisopropylamide, LDA, pKa 36). Ether solvents like tetrahydrofuran (THF) are commonly used for enolate anion formation. With the exception of sodium hydride and sodium amide, most of these bases are soluble in THF. Certain other strong bases, such as alkyl lithium and Grignard reagents, cannot be used to make enolate anions because they rapidly and irreversibly add to carbonyl groups. Nevertheless, these very strong bases are useful in making soluble amide bases. In the preparation of lithium diisopropylamide (LDA), for example, the only other product is the gaseous alkane butane.
Because of its solubility in THF, LDA is a widely used base for enolate anion formation. In this application, one equivalent of diisopropylamine is produced along with the lithium enolate, but this normally does not interfere with the enolate reactions and is easily removed from the products by washing with aqueous acid. Although the reaction of carbonyl compounds with sodium hydride is heterogeneous and slow, sodium enolates are formed with the loss of hydrogen, and no other organic compounds are produced.
The presence of these overlapping p orbitals gives $\alpha$ hydrogens (Hydrogens on carbons adjacent to carbonyls) special properties. In particular, $\alpha$ hydrogens are weakly acidic because the conjugate base, called an enolate, is stabilized though conjugation with the $\pi$ orbitals of the carbonyl. The effect of the carbonyl is seen when comparing the pKa for the $\alpha$ hydrogens of aldehydes (~16-18), ketones (~19-21), and esters (~23-25) to the pKa of an alkane (~50).
Of the two resonance structures of the enolate ion the one which places the negative charge on the oxygen is the most stable. This is because the negative change will be better stabilized by the greater electronegativity of the oxygen.
Acidity of α-Hydrogens in Some Activated Compounds
Compound RCH2–NO2 RCH2–COR RCH2–C≡N RCH2–SO2R
pKa 9 20 25 25
Examples
If the formed enolate is stabilized by more than one carbonyl it is possible to use a weaker base such as sodium ethoxide.
NaOCH2CH3 = Na+ -OCH2CH3 = NaOEt
Because of the acidity of α hydrogens, carbonyls undergo keto-enol tautomerism. Tautomers are rapidly interconverted constitutional isomers, usually distinguished by a different bonding location for a labile hydrogen atom and a differently located double bond. The equilibrium between tautomers is not only rapid under normal conditions, but it often strongly favors one of the isomers (acetone, for example, is 99.999% keto tautomer). Even in such one-sided equilibria, evidence for the presence of the minor tautomer comes from the chemical behavior of the compound. Tautomeric equilibria are catalyzed by traces of acids or bases that are generally present in most chemical samples.
The Ambident Character of Enolate Anions
Since the negative charge of an enolate anion is delocalized over the alpha-carbon and the oxygen, as shown earlier, electrophiles may bond to either atom. Reactants having two or more reactive sites are called ambident, so this term is properly applied to enolate anions. Modestly electrophilic reactants such as alkyl halides are not sufficiently reactive to combine with neutral enol tautomers, but the increased nucleophilicity of the enolate anion conjugate base permits such reactions to take place. Because alkylations are usually irreversible, their products should reflect the inherent (kinetic) reactivity of the different nucleophilic sites.
If an alkyl halide undergoes an SN2 reaction at the carbon atom of an enolate anion the product is an alkylated aldehyde or ketone. On the other hand, if the SN2 reaction occurs at oxygen the product is an ether derivative of the enol tautomer; such compounds are stable in the absence of acid and may be isolated and characterized. These alkylations (shown above) are irreversible under the conditions normally used for SN2 reactions, so the product composition should provide a measure of the relative rates of substitution at carbon versus oxygen. It has been found that this competition is sensitive to a number of factors, including negative charge density, solvation, cation coordination and product stability.
Enolate of Unsymmetrical Carbonyl Compounds
Now let’s consider what happens when an unsymmetrical carbonyl is treated with a base. In the case displayed below there are two possible enolates which can form. The removal of the 2o hydrogen forms the kinetic enolate and is formed faster because it is less substituted and thereby less sterically hindered. The removal of the 3o hydrogen forms the thermodynamic enolate which is more stable because it is more substituted.
Kinetic Enolates
Kinetic enolates are formed when a strong bulky base like LDA is used. The bulky base finds the 2o hydrogen less sterically hindered and preferable removes it.
Low temperature are typically used when forming the kinetic enolate to prevent equilibration to the more stable thermodynamic enolate. Typically a temperature of -78 oC is used.
Thermodynamic Enolates
The thermodynamic enolate is favored by conditions which allow for equilibration. The thermodynamic enolate is usually formed by using a strong base at room temperature. At equilibrium the lower energy of the thermodynamic enolate is preferred, so that the more stable, more stubstituted enolate is formed. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/18%3A_Enols_Enolates_and_the_Aldol_Condensation%3A_ab-Unsaturated_Aldehydes_and_Ketones/18.01%3A_Acidity__of_Aldehydes_and__Ketones%3A_Enolate__Ions.txt |
Acidity of Alpha Hydrogens
Functional groups, such as aldehydes, ketones and esters, contain a carbonyl group which is made up of a sp2 hybridized carbon and oxygen. Because they are sp2 hybridized the carbon and oxygen both have unhybridized p orbitals which can overlap to form the C=O $\pi$ bond.
Keto-enol Tautomerism
Because of the acidity of α hydrogens carbonyls undergo keto-enol tautomerism. Tautomers are rapidly interconverted constitutional isomers, usually distinguished by a different bonding location for a labile hydrogen atom and a differently located double bond. The equilibrium between tautomers is not only rapid under normal conditions, but it often strongly favors one of the isomers (acetone, for example, is 99.999% keto tautomer). Even in such one-sided equilibria, evidence for the presence of the minor tautomer comes from the chemical behavior of the compound. Tautomeric equilibria are catalyzed by traces of acids or bases that are generally present in most chemical samples.
Mechanism for Enol Formation
Acid conditions
1) Protonation of the Carbonyl
2) Enol formation
Basic conditions
1) Enolate formation
2) Protonation
Contributors
• William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry
• Prof. Steven Farmer (Sonoma State University)
Due to the acidic nature of α hydrogens they can be exchanged with deuterium by reaction with D2O (heavy water). The process is accelerated by the addition of an acid or base; an excess of D2O is required. The end result is the complete exchange of all α hydrogens with deuterium.
Mechanism in basic conditions
Step 1: Enolate Formation
Step 2: Deuteration
Exercise $1$
Please draw the product for the following reactions.
Answer
18.03: Halogenation of Aldehydes and Ketones
A carbonyl containing compound with α hydrogens can undergo a substitution reaction with halogens. This reaction comes about because of the tendency of carbonyl compounds to form enolates in basic condition and enols in acidic condition. In these cases even weak bases, such as the hydroxide anion, is sufficient enough to cause the reaction to occur because it is not necessary for a complete conversion to the enolate. For this reaction Cl2, Br2 or I2 can be used as the halogens.
General reaction
Example
Acid Catalyzed Mechanism
Under acidic conditions the reaction occurs thought the formation of an enol which then reacts with the halogen.
1) Protonation of the carbonyl
2) Enol formation
3) SN2 attack
4) Deprotonation
Base Catalyzed Mechanism
Under basic conditions the enolate forms and then reacts with the halogen. Note! This is base promoted and not base catalyzed because an entire equivalent of base is required.
1) Enolate formation
2) SN2 attack
Overreaction during base promoted α halogenation
The fact that an electronegative halogen is placed on an α carbon means that the product of a base promoted α halogenation is actually more reactive than the starting material. The electron withdrawing effect of the halogen makes the α carbon even more acidic and therefor promotes further reaction. Because of this multiple halogenations can occur. This effect is exploited in the haloform reaction discussed later. If a monohalo product is required then acidic conditions are usually used.
The Haloform Reaction
Methyl ketones typically undergo halogenation three times to give a trihalo ketone due to the increased reactivity of the halogenated product as discussed above. This trihalomethyl group is an effective leaving group due to the three electron withdrawing halogens and can be cleaved by a hydroxide anion to effect the haloform reaction. The product of this reaction is a carboxylate and a haloform molecule (CHCl3, CHBr3, CHI3). Overall the haloform reaction represents an effective method for the conversion of methyl ketones to carboxylic acids. Typically, this reaction is performed using iodine because the subsequent iodoform (CHI3) is a bright yellow solid which is easily filtered off.
General reaction
Example: The Haloform Reaction
Mechanism
1) Formation of the trihalo species
2) Nulceophilic attack on the carbonyl carbon
3) Removal of the leaving group
4) Deprotonation
Problems
1) Please draw the products of the following reactions
Answers
1)
Contributors
• Prof. Steven Farmer (Sonoma State University)
18.04: Alkylation of Aldehydes and Ketones
Enolates can act as a nucleophile in SN2 type reactions. Overall an α hydrogen is replaced with an alkyl group. This reaction is one of the more important for enolates because a carbon-carbon bond is formed. These alkylations are affected by the same limitations as SN2 reactions previously discussed. A good leaving group, Chloride, Bromide, Iodide, Tosylate, should be used. Also, secondary and tertiary leaving groups should not be used because of poor reactivity and possible competition with elimination reactions. Lastly, it is important to use a strong base, such as LDA or sodium amide, for this reaction. Using a weaker base such as hydroxide or an alkoxide leaves the possibility of multiple alkylation’s occurring.
Example 1: Alpha Alkylation
Mechanism
1) Enolate formation
2) Sn2 attack
Alkylation of Unsymmetrical Ketones
Unsymmetrical ketones can be regioselctively alkylated to form one major product depending on the reagents.
Treatment with LDA in THF at -78oC tends to form the less substituted kinetic enolate.
Using sodium ethoxide in ethanol at room temperature forms the more substituted thermodynamic enolate.
Problems
1) Please write the structure of the product for the following reactions.
Answers
1) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/18%3A_Enols_Enolates_and_the_Aldol_Condensation%3A_ab-Unsaturated_Aldehydes_and_Ketones/18.02%3A_Keto-Enol_Equilibria.txt |
A useful carbon-carbon bond-forming reaction known as the Aldol Reaction is yet another example of electrophilic substitution at the alpha carbon in enolate anions. The fundamental transformation in this reaction is a dimerization of an aldehyde (or ketone) to a beta-hydroxy aldehyde (or ketone) by alpha C–H addition of one reactant molecule to the carbonyl group of a second reactant molecule. Due to the carbanion like nature of enolates they can add to carbonyls in a similar manner as Grignard reagents. For this reaction to occur at least one of the reactants must have α hydrogens.
Going from reactants to products simply
Example 1: Aldol Reactions
Aldol Reaction Mechanism
Step 1: Enolate formation
Step 2: Nucleophilic attack by the enolate
Step 3: Protonation
Aldol Condensation: the dehydration of Aldol products to synthesize α, β unsaturated carbonyl (enones)
The products of aldol reactions often undergo a subsequent elimination of water, made up of an alpha-hydrogen and the beta-hydroxyl group. The product of this $\beta$-elimination reaction is an α,β-unsaturated aldehyde or ketone. Base-catalyzed elimination occurs with heating. The additional stability provided by the conjugated carbonyl system of the product makes some aldol reactions thermodynamically and mixtures of stereoisomers (E & Z) are obtained from some reactions. Reactions in which a larger molecule is formed from smaller components, with the elimination of a very small by-product such as water, are termed Condensations. Hence, the following examples are properly referred to as aldol condensations. Overall the general reaction involves a dehydration of an aldol product to form an alkene:
Figure: General reaction for an aldol condensation
Going from reactants to products simply
Figure: The aldol condensatio example
Example 2: Aldol Condensation
Aldol Condensation Mechanism
1) Form enolate
2) Form enone
When performing both reactions together always consider the aldol product first then convert to the enone. Note! The double bond always forms in conjugation with the carbonyl.
Example
18.06: Crossed Aldol Condensation
Mixed Aldol Reaction and Condensations
The previous examples of aldol reactions and condensations used a common reactant as both the enolic donor and the electrophilic acceptor. The product in such cases is always a dimer of the reactant carbonyl compound. Aldol condensations between different carbonyl reactants are called crossed or mixed reactions, and under certain conditions such crossed aldol condensations can be effective.
Example 3: Mixed Aldol Reaction
The success of these mixed aldol reactions is due to two factors. First, aldehydes are more reactive acceptor electrophiles than ketones, and formaldehyde is more reactive than other aldehydes. Second, aldehydes lacking alpha-hydrogens can only function as acceptor reactants, and this reduces the number of possible products by half. Mixed aldols in which both reactants can serve as donors and acceptors generally give complex mixtures of both dimeric (homo) aldols and crossed aldols. Because of this most mixed aldol reactions are usually not performed unless one reactant has no alpha hydrogens.
The following abbreviated formulas illustrate the possible products in such a case, red letters representing the acceptor component and blue the donor. If all the reactions occurred at the same rate, equal quantities of the four products would be obtained. Separation and purification of the components of such a mixture would be difficult.
AACH2CHO + BCH2CHO + NaOH → AA + BB + AB + BA
The aldol condensation of ketones with aryl aldehydes to form α,β-unsaturated derivatives is called the Claisen-Schmidt reaction.
Example 4: Claisen-Schmidt Reaction
18.07: Keys to Success: Competitive Rection Pathways and the Int
Intramolecular aldol reaction
Molecules which contain two carbonyl functionalities have the possibility of forming a ring through an intramolecular aldol reaction. In most cases two sets of $\alpha$ hydrogens need to be considered. As with most ring forming reaction five and six membered rings are preferred.
As with other aldol reaction the addition of heat causes an aldol condensation to occur. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/18%3A_Enols_Enolates_and_the_Aldol_Condensation%3A_ab-Unsaturated_Aldehydes_and_Ketones/18.05%3A_Attack__by_Enolates_on_the_Carbonyl___Function%3A_Aldol__Co.txt |
One of the largest and most diverse classes of reactions is composed of nucleophilic additions to a carbonyl group. Conjugation of a double bond to a carbonyl group transmits the electrophilic character of the carbonyl carbon to the beta-carbon of the double bond. These conjugated carbonyl are called enones or α, β unsaturated carbonyls. A resonance description of this transmission is shown below.
From this formula it should be clear that nucleophiles may attack either at the carbonyl carbon, as for any aldehyde, ketone or carboxylic acid derivative, or at the beta-carbon. These two modes of reaction are referred to as 1,2-addition and 1,4-addition respectively. A 1,4-addition is also called a conjugate addition.
Basic reaction of 1,2 addition
Here the nucleophile adds to the carbon which is in the one position. The hydrogen adds to the oxygen which is in the two position.
Basic reaction of 1,4 addition
In 1,4 addition the Nucleophile is added to the carbon β to the carbonyl while the hydrogen is added to the carbon α to the carbonyl.
Mechanism for 1,4 addition
1) Nucleophilic attack on the carbon β to the carbonyl
2) Proton Transfer
Here we can see why this addition is called 1,4. The nucleophile bonds to the carbon in the one position and the hydrogen adds to the oxygen in the four position.
3) Tautomerization
Going from reactant to products simplified
1,2 vs. 1,4 addition
Whether 1,2 or 1,4-addition occurs depends on multiple variables but mostly it is determined by the nature of the nucleophile. During the addition of a nucleophile there is a competition between 1,2 and 1,4 addition products. If the nucleophile is a strong base, such as Grignard reagents, both the 1,2 and 1,4 reactions are irreversible and therefor are under kinetic control. Since 1,2-additions to the carbonyl group are fast, we would expect to find a predominance of 1,2-products from these reactions.
If the nucleophile is a weak base, such as alcohols or amines, then the 1,2 addition is usually reversible. This means the competition between 1,2 and 1,4 addition is under thermodynamic control. In this case 1,4-addition dominates because the stable carbonyl group is retained.
Water
Alcohols
Thiols
1o Amines
2o Amines
HBr
Cyanides
Gilman Reagents
Another important reaction exhibited by organometallic reagents is metal exchange. Organolithium reagents react with cuprous iodide to give a lithium dimethylcopper reagent, which is referred to as a Gilman reagent. Gilman reagents are a source of carbanion like nucleophiles similar to Grignard and Organo lithium reagents. However, the reactivity of organocuprate reagents is slightly different and this difference will be exploited in different situations. In the case of α, β unsaturated carbonyls organocuprate reagents allow for an 1,4 addition of an alkyl group. As we will see later Grignard and Organolithium reagents add alkyl groups 1,2 to α, β unsaturated carbonyls
Organocuprate reagents are made from the reaction of organolithium reagents and $CuI$
$2 RLi + CuI \rightarrow R_2CuLi + LiI$
This acts as a source of R:-
$2 CH_3Li + CuI \rightarrow (CH_3)_2CuLi + LiI$
Example
Nucleophiles which add 1,2 to α, β unsaturated carbonyls
Metal Hydrides
Grignard Reagents
Organolithium Reagents
Contributors
Prof. Steven Farmer (Sonoma State University)
William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/18%3A_Enols_Enolates_and_the_Aldol_Condensation%3A_ab-Unsaturated_Aldehydes_and_Ketones/18.09%3A_Conjugate_Additions_to_ab-Unsaturated_Aldehydes___and__Keto.txt |
One of the largest and most diverse classes of reactions is composed of nucleophilic additions to a carbonyl group. Conjugation of a double bond to a carbonyl group transmits the electrophilic character of the carbonyl carbon to the beta-carbon of the double bond. These conjugated carbonyl are called enones or α, β unsaturated carbonyls. A resonance description of this transmission is shown below.
From this formula it should be clear that nucleophiles may attack either at the carbonyl carbon, as for any aldehyde, ketone or carboxylic acid derivative, or at the beta-carbon. These two modes of reaction are referred to as 1,2-addition and 1,4-addition respectively. A 1,4-addition is also called a conjugate addition.
Basic reaction of 1,2 addition
Here the nucleophile adds to the carbon which is in the one position. The hydrogen adds to the oxygen which is in the two position.
Basic reaction of 1,4 addition
In 1,4 addition the Nucleophile is added to the carbon β to the carbonyl while the hydrogen is added to the carbon α to the carbonyl.
Mechanism for 1,4 addition
1) Nucleophilic attack on the carbon β to the carbonyl
2) Proton Transfer
Here we can see why this addition is called 1,4. The nucleophile bonds to the carbon in the one position and the hydrogen adds to the oxygen in the four position.
3) Tautomerization
Going from reactant to products simplified
1,2 vs. 1,4 addition
Whether 1,2 or 1,4-addition occurs depends on multiple variables but mostly it is determined by the nature of the nucleophile. During the addition of a nucleophile there is a competition between 1,2 and 1,4 addition products. If the nucleophile is a strong base, such as Grignard reagents, both the 1,2 and 1,4 reactions are irreversible and therefor are under kinetic control. Since 1,2-additions to the carbonyl group are fast, we would expect to find a predominance of 1,2-products from these reactions.
If the nucleophile is a weak base, such as alcohols or amines, then the 1,2 addition is usually reversible. This means the competition between 1,2 and 1,4 addition is under thermodynamic control. In this case 1,4-addition dominates because the stable carbonyl group is retained.
Water
Alcohols
Thiols
1o Amines
2o Amines
HBr
Cyanides
Gilman Reagents
Another important reaction exhibited by organometallic reagents is metal exchange. Organolithium reagents react with cuprous iodide to give a lithium dimethylcopper reagent, which is referred to as a Gilman reagent. Gilman reagents are a source of carbanion like nucleophiles similar to Grignard and Organo lithium reagents. However, the reactivity of organocuprate reagents is slightly different and this difference will be exploited in different situations. In the case of α, β unsaturated carbonyls organocuprate reagents allow for an 1,4 addition of an alkyl group. As we will see later Grignard and Organolithium reagents add alkyl groups 1,2 to α, β unsaturated carbonyls
Organocuprate reagents are made from the reaction of organolithium reagents and $CuI$
$2 RLi + CuI \rightarrow R_2CuLi + LiI$
This acts as a source of R:-
$2 CH_3Li + CuI \rightarrow (CH_3)_2CuLi + LiI$
Example
Nucleophiles which add 1,2 to α, β unsaturated carbonyls
Metal Hydrides
Grignard Reagents
Organolithium Reagents
Contributors
Prof. Steven Farmer (Sonoma State University)
William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry
18.11: Conjugate Additions of Enolate Ions: Michael Addition
Basic reaction of 1,4 addition
In 1,4 addition the Nucleophile is added to the carbon β to the carbonyl while the hydrogen is added to the carbon α to the carbonyl.
Mechanism for 1,4 addition
1) Nucleophilic attack on the carbon β to the carbonyl
2) Proton Transfer
Here we can see why this addition is called 1,4. The nucleophile bonds to the carbon in the one position and the hydrogen adds to the oxygen in the four position.
3) Tautomerization
Enolates undergo 1,4 addition to α, β-unsaturated carbonyl compounds is a process called a Michael addition. The reaction is named after American chemist Arthur Michael (1853-1942).
Contributors
Many times the product of a Michael addition produces a dicarbonyl which can then undergo an intramolecular aldol reaction. These two processes together in one reaction creates two new carbon-carbon bonds and also creates a ring. Ring-forming reactions are called annulations after the Latin work for ring annulus. The reaction is named after English chemist Sir Robert Robinson (1886-1975) who developed it. He received the Nobel prize in chemistry in 1947. Remember that during annulations five and six membered rings are preferred. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/18%3A_Enols_Enolates_and_the_Aldol_Condensation%3A_ab-Unsaturated_Aldehydes_and_Ketones/18.10%3A_12-_and__14-Additions_of_Organometallic_Reagents.txt |
The IUPAC system of nomenclature assigns a characteristic suffix to these classes. The –e ending is removed from the name of the parent chain and is replaced -anoic acid. Since a carboxylic acid group must always lie at the end of a carbon chain, it is always is given the #1 location position in numbering and it is not necessary to include it in the name.
Many carboxylic acids are called by the common names. These names were chosen by chemists to usually describe a source of where the compound is found. In common names of aldehydes, carbon atoms near the carboxyl group are often designated by Greek letters. The atom adjacent to the carbonyl function is alpha, the next removed is beta and so on.
Formula
Common Name
Source
IUPAC Name
Melting Point
Boiling Point
HCO2H formic acid ants (L. formica) methanoic acid 8.4 ºC 101 ºC
CH3CO2H acetic acid vinegar (L. acetum) ethanoic acid 16.6 ºC 118 ºC
CH3CH2CO2H propionic acid milk (Gk. protus prion) propanoic acid -20.8 ºC 141 ºC
CH3(CH2)2CO2H butyric acid butter (L. butyrum) butanoic acid -5.5 ºC 164 ºC
CH3(CH2)3CO2H valeric acid valerian root pentanoic acid -34.5 ºC 186 ºC
CH3(CH2)4CO2H caproic acid goats (L. caper) hexanoic acid -4.0 ºC 205 ºC
CH3(CH2)5CO2H enanthic acid vines (Gk. oenanthe) heptanoic acid -7.5 ºC 223 ºC
CH3(CH2)6CO2H caprylic acid goats (L. caper) octanoic acid 16.3 ºC 239 ºC
CH3(CH2)7CO2H pelargonic acid pelargonium (an herb) nonanoic acid 12.0 ºC 253 ºC
CH3(CH2)8CO2H capric acid goats (L. caper) decanoic acid 31.0 ºC 219 ºC
Example (Common Names Are in Red)
Naming carboxyl groups added to a ring
When a carboxyl group is added to a ring the suffix -carboxylic acid is added to the name of the cyclic compound. The ring carbon attached to the carboxyl group is given the #1 location number.
Naming carboxylates
Salts of carboxylic acids are named by writing the name of the cation followed by the name of the acid with the –ic acid ending replaced by an –ate ending. This is true for both the IUPAC and Common nomenclature systems.
Naming carboxylic acids which contain other functional groups
Carboxylic acids are given the highest nomenclature priority by the IUPAC system. This means that the carboxyl group is given the lowest possible location number and the appropriate nomenclature suffix is included. In the case of molecules containing carboxylic acid and alcohol functional groups the OH is named as a hydroxyl substituent. However, the l in hydroxyl is generally removed.
In the case of molecules containing a carboxylic acid and aldehydes and/or ketones functional groups the carbonyl is named as a "Oxo" substituent.
In the case of molecules containing a carboxylic acid an amine functional group the amine is named as an "amino" substituent.
When carboxylic acids are included with an alkene the following order is followed:
(Location number of the alkene)-(Prefix name for the longest carbon chain minus the -ane ending)-(an –enoic acid ending to indicate the presence of an alkene and carboxylic acid)
Remember that the carboxylic acid has priority so it should get the lowest possible location number. Also, remember that cis/tran or E/Z nomenclature for the alkene needs to be included if necessary.
Naming dicarboxylic acids
For dicarboxylic acids the location numbers for both carboxyl groups are omitted because both functional groups are expected to occupy the ends of the parent chain. The ending –dioic acid is added to the end of the parent chain. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/19%3A_Carboxylic_Acids/19.01%3A_Naming__the_Carboxylic_Acids.txt |
Structure of the carboxyl acid group
Carboxylic acids are organic compounds which incorporate a carboxyl functional group, CO2H. The name carboxyl comes from the fact that a carbonyl and a hydroxyl group are attached to the same carbon.
The carbon and oxygen in the carbonyl are both sp2 hybridized which give a carbonyl group a basic trigonal shape. The hydroxyl oxygen is also sp2 hybridized which allows one of its lone pair electrons to conjugate with the pi system of the carbonyl group. This make the carboxyl group planar an can represented with the following resonance structure.
Carboxylic acids are named such because they can donate a hydrogen to produce a carboxylate ion. The factors which affect the acidity of carboxylic acids will be discussed later.
Physical Properties of Some Carboxylic Acids
The table at the beginning of this page gave the melting and boiling points for a homologous group of carboxylic acids having from one to ten carbon atoms. The boiling points increased with size in a regular manner, but the melting points did not. Unbranched acids made up of an even number of carbon atoms have melting points higher than the odd numbered homologs having one more or one less carbon. This reflects differences in intermolecular attractive forces in the crystalline state. In the table of fatty acids we see that the presence of a cis-double bond significantly lowers the melting point of a compound. Thus, palmitoleic acid melts over 60º lower than palmitic acid, and similar decreases occur for the C18 and C20 compounds. Again, changes in crystal packing and intermolecular forces are responsible.
The factors that influence the relative boiling points and water solubilities of various types of compounds were discussed earlier. In general, dipolar attractive forces between molecules act to increase the boiling point of a given compound, with hydrogen bonds being an extreme example. Hydrogen bonding is also a major factor in the water solubility of covalent compounds To refresh your understanding of these principles Click Here. The following table lists a few examples of these properties for some similar sized polar compounds (the non-polar hydrocarbon hexane is provided for comparison).
Physical Properties of Some Organic Compounds
Formula
IUPAC Name
Molecular Weight
Boiling Point
Water Solubility
CH3(CH2)2CO2H butanoic acid 88 164 ºC very soluble
CH3(CH2)4OH 1-pentanol 88 138 ºC slightly soluble
CH3(CH2)3CHO pentanal 86 103 ºC slightly soluble
CH3CO2C2H5 ethyl ethanoate 88 77 ºC moderately soluble
CH3CH2CO2CH3 methyl propanoate 88 80 ºC slightly soluble
CH3(CH2)2CONH2 butanamide 87 216 ºC soluble
CH3CON(CH3)2 N,N-dimethylethanamide 87 165 ºC very soluble
CH3(CH2)4NH2 1-aminobutane 87 103 ºC very soluble
CH3(CH2)3CN pentanenitrile 83 140 ºC slightly soluble
CH3(CH2)4CH3 hexane 86 69 ºC insoluble
The first five entries all have oxygen functional groups, and the relatively high boiling points of the first two is clearly due to hydrogen bonding. Carboxylic acids have exceptionally high boiling points, due in large part to dimeric associations involving two hydrogen bonds. A structural formula for the dimer of acetic acid is shown here. When the mouse pointer passes over the drawing, an electron cloud diagram will appear. The high boiling points of the amides and nitriles are due in large part to strong dipole attractions, supplemented in some cases by hydrogen bonding.
19.03: Spectroscopy and Mass Spectrometry of Carboxylic Acids
IR
The carboxyl group is associated with two characteristic infrared stretching absorptions which change markedly with hydrogen bonding. The spectrum of a CCl4 solution of propionic acid (propanoic acid), shown below, is illustrative. Carboxylic acids exist predominantly as hydrogen bonded dimers in condensed phases. The O-H stretching absorption for such dimers is very strong and broad, extending from 2500 to 3300 cm-1. This absorption overlaps the sharper C-H stretching peaks, which may be seen extending beyond the O-H envelope at 2990, 2950 and 2870 cm-1. The smaller peaks protruding near 2655 and 2560 are characteristic of the dimer. In ether solvents a sharper hydrogen bonded monomer absorption near 3500 cm-1 is observed, due to competition of the ether oxygen as a hydrogen bond acceptor. The carbonyl stretching frequency of the dimer is found near 1710 cm-1, but is increased by 25 cm-1 or more in the monomeric state. Other characteristic stretching and bending absorptions are marked in the spectrum.
NMR
The combination of anisotropy and electronegativity causes the O-H hydrogen in a carboxylic acid to be very deshielded.
Hydrogen environments adjacent to a carboxylic acid are shifted to the region of 2.5-3.0 ppm.Deshielding occurs due to the fact that the sp2 hybridized carbon the the carboxylic acid is more electronegative than a sp3 hybridized carbon. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/19%3A_Carboxylic_Acids/19.02%3A_Structural_and_Physical__Properties_of_Carboxylic_Acids.txt |
Objectives
After completing this section, you should be able to
1. identify the form that carboxylic acids take within living cells.
2. use the Henderson‑Hasselbalch equation to calculate the percentages of dissociated and undissociated acids [A] and [HA] in a solution, given the pKa value of the acid and the pH of the solution.
3. explain why cellular carboxylic acids are always referred to by the name of their anion.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• Henderson‑Hasselbalch equation
• physiological pH
Organic molecules in buffered solution: The Henderson-Hasselbalch equation
The environment inside a living cell, where most biochemical reactions take place, is an aqueous buffer with pH ~ 7 (or sometimes referred to as the physiological pH). Recall from your General Chemistry course that a buffer is a solution of a weak acid and its conjugate base. The key equation for working with buffers is the Henderson-Hasselbalch equation:
The Henderson-Hasselbalch equation:
The equation tells us that if our buffer is an equimolar solution of a weak acid and its conjugate base, the pH of the buffer will equal the pKa of the acid (because the log of 1 is equal to zero). If there is more of the acid form than the base, then of course the pH of the buffer is lower than the pKa of the acid.
Exercise 7.2.4
What is the pH of an aqueous buffer solution that is 30 mM in acetic acid and 40 mM in sodium acetate? The pKa of acetic acid is 4.8.
Solution
The Henderson-Hasselbalch equation is particularly useful when we want to think about the protonation state of different biomolecule functional groups in a pH 7 buffer. When we do this, we are always assuming that the concentration of the biomolecule is small compared to the concentration of the buffer components. (The actual composition of physiological buffer is complex, but it is primarily based on phosphoric and carbonic acids.).
Imagine an aspartic acid residue located on the surface of a protein in a human cell. Being on the surface, the side chain is in full contact with the pH 7 buffer surrounding the protein. In what state is the side chain functional group: the protonated state (a carboxylic acid) or the deprotonated state (a carboxylate ion)? Using the Henderson-Hasselbalch equation, we fill in our values for the pH of the buffer and a rough pKa approximation of pKa = 5 for the carboxylic acid functional group. Doing the math, we find that the ratio of carboxylate to carboxylic acid is about 100 to 1: the carboxylic acid is almost completely ionized (in the deprotonated state) inside the cell. This result extends to all other carboxylic acid groups you might find on natural biomolecules or drug molecules: in the physiological environment, carboxylic acids are almost completely deprotonated. Indeed, often cellular carboxylic acids are often referred to by the name of their anion. So rather than pyruvic acid, or acetic acid or lactic acid, it is pyruvate, acetate or lactate.
While we are most interested in the state of molecules at pH 7, the Henderson-Hasselbalch equation can of course be used to determine the protonation state of functional groups in solutions buffered to other pH levels. The exercise below provide some practice in this type of calculation.
Exercise 7.2.6
What is the ratio of acetate ion to neutral acetic acid when a small amount of acetic acid (pKa = 4.8) is dissolved in a buffer of pH 2.8? pH 3.8? pH 4.8? pH 5.8? pH 6.8?
Solution
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris)
The resonance effect described here is undoubtedly the major contributor to the exceptional acidity of carboxylic acids. However, inductive effects also play a role. For example, alcohols have pKa's of 16 or greater but their acidity is increased by electron withdrawing substituents on the alkyl group. The following diagram illustrates this factor for several simple inorganic and organic compounds (row #1), and shows how inductive electron withdrawal may also increase the acidity of carboxylic acids (rows #2 & 3). The acidic hydrogen is colored red in all examples.
Water is less acidic than hydrogen peroxide because hydrogen is less electronegative than oxygen, and the covalent bond joining these atoms is polarized in the manner shown. Alcohols are slightly less acidic than water, due to the poor electronegativity of carbon, but chloral hydrate, Cl3CCH(OH)2, and 2,2,2,-trifluoroethanol are significantly more acidic than water, due to inductive electron withdrawal by the electronegative halogens (and the second oxygen in chloral hydrate). In the case of carboxylic acids, if the electrophilic character of the carbonyl carbon is decreased the acidity of the carboxylic acid will also decrease. Similarly, an increase in its electrophilicity will increase the acidity of the acid. Acetic acid is ten times weaker an acid than formic acid (first two entries in the second row), confirming the electron donating character of an alkyl group relative to hydrogen, as noted earlier in a discussion of carbocation stability. Electronegative substituents increase acidity by inductive electron withdrawal. As expected, the higher the electronegativity of the substituent the greater the increase in acidity (F > Cl > Br > I), and the closer the substituent is to the carboxyl group the greater is its effect (isomers in the 3rd row). Substituents also influence the acidity of benzoic acid derivatives, but resonance effects compete with inductive effects. The methoxy group is electron donating and the nitro group is electron withdrawing (last three entries in the table of pKa values).
For additional information about substituent effects on the acidity of carboxylic acids Click Here | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/19%3A_Carboxylic_Acids/19.04%3A_Acidic__and_Basic__Character_of_Carboxylic_Acids.txt |
Objectives
After completing this section, you should be able to
1. describe in detail the methods of preparing carboxylic acids discussed in previous chapters:
1. the oxidation of alkylbenzenes.
2. the oxidative cleavage of alkenes.
3. the oxidation of primary alcohols and aldehydes.
2. discuss, in detail, the hydrolysis of nitriles:
1. write an equation to illustrate the preparation of a carboxylic acid through nucleophilic attack by cyanide ion on an alkyl halide and hydrolysis of the nitrile which results.
2. identify the carboxylic acid formed from the hydrolysis of a given nitrile, or from the reaction of a given alkyl halide with cyanide ion followed by hydrolysis of the resulting nitrile.
3. identify the alkyl halide needed to prepare a given carboxylic acid by the formation and subsequent hydrolysis of a nitrile.
4. identify the reagents needed to convert a given alkyl halide into a carboxylic acid containing one more carbon atom.
3. discuss, in detail, the carboxylation of Grignard reagents:
1. write an equation describing the formation of a carboxylic acid from a Grignard reagent.
2. identify the carboxylic acid obtained through the treatment of a given Grignard reagent with carbon dioxide followed by dilute acid.
3. identify the Grignard reagent (or the alkyl halide required to form the Grignard reagent) that must be used to produce a given carboxylic acid by reaction with carbon dioxide.
4. write the detailed mechanism for the formation of a carboxylic acid using a Grignard reagent.
Study Notes
Review the methods of obtaining carboxylic acids presented in earlier sections:
1. oxidation of aromatic compounds—Section 16.9.
2. oxidative cleavage of alkenes—Section 8.8.
3. oxidation of primary alcohols and aldehydes—Sections 17.7 and 19.3.
The carbon atom of a carboxyl group has a high oxidation state. It is not surprising, therefore, that many of the chemical reactions used for their preparation are oxidations. Such reactions have been discussed in previous sections of this text, and the following diagram summarizes most of these. To review the previous discussion of any of these reaction classes simply click on the number (1 to 4) or descriptive heading for the group.
Two other useful procedures for preparing carboxylic acids involve hydrolysis of nitriles and carboxylation of organometallic intermediates. As shown in the following diagram, both methods begin with an organic halogen compound and the carboxyl group eventually replaces the halogen. Both methods require two steps, but are complementary in that the nitrile intermediate in the first procedure is generated by a SN2 reaction, in which cyanide anion is a nucleophilic precursor of the carboxyl group. The hydrolysis may be either acid or base-catalyzed, but the latter give a carboxylate salt as the initial product.
In the second procedure the electrophilic halide is first transformed into a strongly nucleophilic metal derivative, and this adds to carbon dioxide (an electrophile). The initial product is a salt of the carboxylic acid, which must then be released by treatment with strong aqueous acid.
An existing carboxylic acid may be elongated by one methylene group, using a homologation procedure called the Arndt-Eistert reaction. To learn about this useful method Click Here. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/19%3A_Carboxylic_Acids/19.06%3A_Methods__for__Introducing_the_Carboxy__Functional__Group.txt |
Objectives
After completing this section, you should be able to identify the four types of reaction which a carboxylic acid can undergo.
Study Notes
You may wish to review Section 17.4 which discusses reduction of carbonyl compounds to form alcohols and Sections 20.2–20.4 which highlights the acidity of carboxylic acids, which is important to salt formation and substitution of the hydroxyl hydrogen. Nucleophilic acyl substitution (Chapter 21) and alpha substitutions (Chapter 22) are discussed later in more detail.
Four Categories of Carboxylic Acid Reactions
The scheme summarizes some of the general reactions that carboxylic acids undergo.
Four general reaction categories are represented here: (1) As carboxylic acid deprotonates quite readily, it is quite easy to form a carboxylate salt or to substitute the hydroxyl hydrogen. (2) The category of nucleophilic acyl substitution represents the substitution of the whole hydroxyl group, which we will see later in more detail leads to several carboxylic acid derivatives (e.g. acid halides, esters, amides, thioesters, acid anhydrides etc.). (3) Like other carbonyl compounds, carboxylic acids can be reduced by reagents like LiAlH4. (4) While the proton on the carbon alpha to the carbonyl group is not as acidic as the hydroxyl hydrogen, it can be removed leading to substitution at the alpha site.
Introduction
Carboxylic acid derivatives are functional groups whose chemistry is closely related. The main difference is the presence of an electronegative substituent that can act as a leaving group during a nucleophile substitution reaction. Although there are many types of carboxylic acid derivatives known, this article focuses on four: acid halides, acid anhydrides, esters, and amides.
General mechanism
1) Nucleophilic attack on the carbonyl
2) Leaving group is removed
Although aldehydes and ketones also contain carbonyls, their chemistry is distinctly different because they do not contain suitable leaving groups. Once a tetrahedral intermediate is formed, aldehydes and ketones cannot reform their carbonyls. Because of this, aldehydes and ketones typically undergo nucleophilic additions and not substitutions.
The relative reactivity of carboxylic acid derivatives toward nucleophile substitutions is related to the electronegative leaving group’s ability to activate the carbonyl. The more electronegative leaving groups withdraw electron density from the carbonyl, thereby increasing its electrophilicity.
19.08: Carboxylic Acid Derivatives: Alkanoyl (Acyl) Halides and Anhydrides
Carboxylic acids react with Thionyl Chloride (\(SOCl_2\)) to form acid chlorides.
During the reaction the hydroxyl group of the carboxylic acid is converted to a chlorosulfite intermediate making it a better leaving group. The chloride anion produced during the reaction acts a nucleophile.
Mechanism
1) Nucleophilic attack on Thionyl Chloride
2) Removal of Cl leaving group
3) Nucleophilic attack on the carbonyl
4) Leaving group removal
5) Deprotonation
Carboxylic acids can react with alcohols to form esters in a process called Fischer esterification.
Usually the alcohol is used as the reaction solvent. An acid catalyst is required.
Mechanism
1) Protonation of the carbonyl by the acid. The carbonyl is now activated toward nucleophilic attack.
2) Nucleophilic attack on the carbonyl
3) Proton transfer
4) Water leaves
5) Deprotonation
Conversion of Carboxylic Acids to Amides
The direct reaction of a carboxylic acid with an amine would be expected to be difficult because the basic amine would deprotonate the carboxylic acid to form a highly unreactive carboxylate. However when the ammonium carboxylate salt is heated to a temperature above 100 oC water is driven off and an amide is formed.
Conversion of Carboxylic acids to amide using DCC as an activating agent
The direct conversion of a carboxylic acid to an amide is difficult because amines are basic and tend to convert carboxylic acids to their highly unreactive carboxylates. In this reaction the carboxylic acid adds to the DCC molecule to form a good leaving group which can then be displaced by an amine during nucleophilic substitution. DCC induced coupling to form an amide linkage is an important reaction in the synthesis of peptides.
Mechanism
1) Deprotonation
2) Nucleophilic attack by the carboxylate
3) Nucleophilic attack by the amine
4) Proton transfer
5) Leaving group removal
19.09: Carboxylic Acid Derivatives: Esters
Fischer esterification is the esterification of a Carboxylic acid by heating it with an alcohol in the presence of a strong acid as the catalyst.
Mechanism
The overall reaction is reversible; to drive the reaction to completion, it is necessary to exploit Le Chateliers principle, which can be done either by continuously removing the water formed from the system or by using a large excess of the alcohol.
1) Protonation of the carbonyl by the acid. The carbonyl is now activated toward nucleophilic attack.
2) Nucleophilic attack on the carbonyl
3) Proton transfer
4) Water leaves
5) Deprotonation | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/19%3A_Carboxylic_Acids/19.07%3A_Substitution_at_the_Carboxy__Carbon%3A_The_Addition-Elimination__Mechanism.txt |
The direct reaction of a carboxylic acid with an amine would be expected to be difficult because the basic amine would deprotonate the carboxylic acid to form a highly unreactive carboxylate. However when the ammonium carboxylate salt is heated to a temperature above 100 oC water is driven off and an amide is formed.
Contributors
Prof. Steven Farmer (Sonoma State University)
19.11: Reduction of Carboxylic Acids by Lithium Aluminum Hydride
Carboxylic acids can be converted to 1o alcohols using Lithium aluminum hydride (LiAlH4). Note that NaBH4 is not strong enough to convert carboxylic acids or esters to alcohols. An aldehyde is produced as an intermediate during this reaction, but it cannot be isolated because it is more reactive than the original carboxylic acid.
Possible Mechanism
1) Deprotonation
2) Nucleophilic attack by the hydride anion
3) Leaving group removal
4) Nucleophilic attack by the hydride anion
5) The alkoxide is protonated
19.12: Bromination Next to the Carboxy Group: The Hell-Volhard-Zelinsky Reaction
Although the alpha bromination of some carbonyl compounds, such as aldehydes and ketones, can be accomplished with Br2 under acidic conditions, the reaction will generally not occur with acids, esters, and amides. This is because only aldehydes and ketones enolize to a sufficient extent to allow the reaction to occur. However, carboxylic acids, can be brominated in the alpha position with a mixture of Br2 and PBr3 in a reaction called the Hell-Volhard-Zelinskii reaction.
The mechanism of this reaction involves an acid bromide enol instead of the expected carboxylic acid enol. The reaction stats with the reaction of the carboxylic acid with PBr3 to form the acid bromide and HBr. The HBr then catalyzes the formation of the acid bromide enol which subsequently reacts with Br2 to give alpha bromination. Lastly, the acid bromide reacts with water to reform the carboxylic acid.
Contributors
Prof. Steven Farmer (Sonoma State University)
19.13: Biological Activity of Carboxylic Acids
Glutamine synthetase
You have already learned that the carboxylate functional group is a very unreactive substrate for an enzyme-catalyzed acyl substitution reactions. How, then, does a living system accomplish an ‘uphill’ reaction such as the one shown below, where glutamate (a carboxylate) is converted to glutamine (an amide)?
It turns out that this conversion is not carried out directly. Rather, the first conversion is from a carboxylate (the least reactive acyl transfer substrate) to an acyl phosphate (the most reactive acyl transfer substrate). This transformation requires a reaction that we are familiar with from chapter 10: phosphorylation of a carboxylate oxygen with ATP as the phosphate donor.
Note that this is just one of the many ways that ATP is used as a energy storage unit: in order to make a high energy acyl phosphate molecule from a low energy carboxylate, the cell must ‘spend’ the energy of one ATP molecule.
The acyl phosphate version of glutamate is now ready to be converted directly to an amide (glutamine) via a nucleophilic acyl substitution reaction, as an ammonia molecule attacks the carbonyl and the phosphate is expelled.
Overall, this reaction can be written as:
Asparagine synthetase
Another common form of activated carboxylate group is an acyl adenosine phosphate. Consider another amino acid reaction, the conversion of aspartate to asparagine. In the first step, the carboxylate group of aspartate must be activated:
Once again, ATP provides the energy for driving the uphill reaction. This time, however, the activated carboxylate takes the form of an acyl adenosine (mono)phosphate. All that has happened is that the carboxylate oxygen has attacked the a-phosphate of ATP rather than the g-phosphate.
The reactive acyl-AMP version of aspartate is now ready to be converted to an amide (asparagine) via nucleophilic attack by ammonia. In the case of glutamine synthase, the source of ammonia was free ammonium ion in solution. In the case of asparagine synthase, the NH3 is derived from the hydrolysis of glutamine (this is simply another acyl substitution reaction):
The hydrolysis reaction is happening in the same enzyme active site – as the NH3 is expelled in the hydrolysis of glutamine, it immediately turns around and acts as the nucleophile in the conversion of aspartyl-AMP to asparagine:
Keep in mind that the same enzyme is also binding ATP and using it to activate aspartate – this is a busy construction zone!
Overall, this reaction can be written in condensed form as:
The use of glutamine as a ‘carrier’ for ammonia is a fairly common strategy in metabolic pathways. This strategy makes sense, as it allows cells to maintain a constant source of NH3 for reactions that require it, without the need for high solution concentrations of free ammonia.
Glycinamide ribonucleotide synthetase
One of the early steps in the construction of purine bases (the adenine and guanine bases in DNA and RNA) involves an acyl substitution reaction with an acyl phosphate intermediate. In this case, the attacking nucleophile is not ammonia but a primary amine. The strategy, however, is similar to that of glutamine synthase. The carboxylate group on glycine is converted to an acyl phosphate, at the cost of one ATP molecule. The acyl group is then transferred to 5-phosphoribosylamine, resulting in an amide product.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/19%3A_Carboxylic_Acids/19.10%3A_Carboxylic_Acid__Derivatives%3A__Amides.txt |
Objectives
After completing this section, you should be able to
1. compare the reactions of carboxylic acid derivatives with nucleophiles to the reactions of aldehydes and ketones with nucleophiles.
2. arrange a given list of carboxylic acid derivatives in order of increasing or decreasing reactivity towards nucleophiles.
3. explain the difference in reactivity towards nucleophiles of two or more given carboxylic acid derivatives.
4. explain why esters and amides are commonly found in nature, but acid halides and acid anhydrides are not.
Study Notes
The general nucleophilic acyl substitution reaction, and its mechanism, were discussed earlier in “III. General Reactions of Carbonyl Compounds.” Review if necessary.
The reading describes the relative reactivities of “biologically relevant acyl groups.” Both acid anhydrides and acid halides readily react with water and cannot exist for any length of time in living organisms. The following scheme illustrates the relative reactivities of most carboxylic acid derivatives that you will encounter.
Carboxylic acid derivatives and acyl groups
The functional groups that undergo nucleophilic acyl substitutions are called carboxylic acid derivatives: these include carboxylic acids themselves, carboxylates (deprotonated carboxylic acids), amides, esters, thioesters, and acyl phosphates. Two more examples of carboxylic acid derivatives which are less biologically relevant but important in laboratory synthesis are carboxylic acid anyhydrides and acid chlorides.
The carboxylic acid derivatives can be distinguished from aldehydes and ketones by the presence of a group containing an electronegative heteroatom - usually oxygen, nitrogen, or sulfur – bonded directly to the carbonyl carbon. You can think of a carboxylic acid derivative as having two sides. One side is the carbonyl group and the attached alkyl group: this is called an acyl group (in the specific case where R is a methyl group, the term acetyl group is used). One the other side is the heteroatom-containing group: in this text, we will sometimes refer to this component as the ‘acyl X' group (this, however, is not a standard term in organic chemistry).
Notice that the acyl X groups are simply deprotonated forms of other functional groups: in an amide, for example, the acyl X group is an amine, while in an ester the acyl X group is an alcohol.
Example 21.2.1
What is the ‘acyl X’ group in:
1. an acid anhydride?
2. a thioester?
3. a carboxylic acid?
Solution
The nucleophilic acyl substitution reaction
The fact that the atom adjacent to the carbonyl carbon in carboxylic acid derivatives is an electronegative heteroatom – rather than a carbon like in ketones or a hydrogen like in aldehydes - is critical to understanding the reactivity of these functional groups. Just like in aldehydes and ketones, carboxylic acid derivatives are attacked from one side of their trigonal planar carbonyl carbon by a nucleophile, converting this carbon to tetrahedral (sp3) geometry. In carboxylic acid derivatives, the acyl X group is a potential leaving group. What this means is that the tetrahedral product formed from attack of the nucleophile on the carbonyl carbon is not the product: it is a reactive intermediate. The tetrahedral intermediate rapidly collapses: the carbon-oxygen double bond re-forms, and the acyl X group is expelled.
Notice that in the product, the nucleophile becomes the new acyl X group. This is why this reaction type is called a nucleophilic acyl substitution: one acyl X group is substituted for another. For example, in the reaction below, one alcohol X group (3-methyl-1-butanol) is replaced by another alcohol X group (methanol), as one ester is converted to another.
Another way of looking at this reaction is to picture the acyl group being transferred from one acyl X group to another: in the example above, the acetyl group is being transferred from 3-methyl-1-butanol to methanol. For this reason, nucleophilic acyl substitutions are also commonly referred to as acyl transfer reactions.
When the incoming nucleophile in an acyl substitution is a water molecule, the reaction is also referred to as an acyl hydrolysis. For example, the following reaction can be described as the hydrolysis of an ester (to form a carboxylic acid and an alcohol).
We could also describe this reaction as the transfer of an acyl group from an alcohol to a water molecule.
In a similar vein, the hydrolysis of an amide to form a carboxylic acid could be described as the transfer of an acyl group from ammonia (NH3) to water.
As we will see in later sections of this chapter the hydrolysis of esters and amides are particularly important reaction types in biochemical pathways. When your body digests the fat in a hamburger, for example, enzymes in your pancreas called lipases first catalyze ester hydrolysis reactions to free the fatty acids (we'll look more closely at this reaction in section 12.4D).
The relative reactivity of carboxylic acid derivatives
The relative reactivity of the carboxylic acid derivatives is an important concept to understand before entering into a detailed examination of nucleophilic acyl substitutions. As a general rule, the carbonyl carbon in an acyl group is less electrophilic than that in an aldehyde or ketone. This is because in carboxylic acid derivatives, the partial positive charge on the carbon is stabilized somewhat by resonance effects from the adjacent heteroatom.
Among the carboxylic acid derivatives, carboxylate groups are the least reactive towards nucleophilic acyl substitution, followed by amides, then esters and (protonated) carboxylic acids, thioesters, and finally acyl phosphates, which are the most reactive among the biologically relevant acyl groups.
The different reactivities of the functional groups can be understood by evaluating the basicity of the leaving group in each case - remember from section 8.5 that weaker bases are better leaving groups! A thioester is more reactive than an ester, for example, because a thiolate (RS-) is a weaker base than an alkoxide (RO-). In general, if the incoming nucleophile is a weaker base than the ‘acyl X’ group that is already there, the first nucleophilic step will simply reverse itself and we’ll get the starting materials back:
This is why it is not possible to directly convert an ester, for example, into a thioester by an acyl substitution reaction – this would be an uphill reaction.
Here’s another way to think about the relative reactivites of the different carboxylic acid derivatives: consider the relative electrophilicity, or degree of partial positive charge, on the carbonyl carbon in each species. This depends on how much electron density the neighboring heteroatom on the acyl X group is able to donate: greater electron donation by the heteroatom implies lower partial positive charge on the carbonyl carbon, which in turn implies lower electrophilicity.
The negatively charged oxygen on the carboxylate group has lots of electron density to donate, thus the carbonyl carbon is not very electrophilic. In amides, the nitrogen atom is a powerful electron donating group by resonance - recall that the carbon-nitrogen bond in peptides has substantial double-bond character - thus amides are relatively unreactive. Amides do undergo acyl substitution reactions in biochemical pathways, but these reactions are inherently slow and the enzymes catalyzing them have evolved efficient strategies to lower the activation energy barrier.
Carboxylic acids and esters are in the middle range of reactivity, while thioesters are somewhat more reactive. The most reactive of the carboxylic acid derivatives frequently found in biomolecules are the acyl phosphates. These are most often present in two forms: the simple acyl monophosphate, and the acyl-adenosine monophosphate.
Both are highly reactive to acyl substitution reactions, and are often referred to as ‘activated acyl groups’ or ‘activated carboxylic acids’. The high reactivity of acyl phosphates is due mainly to the ability to form complexes with magnesium ions.
The magnesium ion acts as a Lewis acid to accept electron density from the oxygen end of the acyl carbonyl bond, thus greatly increasing the degree of partial positive charge - and thus the electrophilicity - of the carbonyl carbon. The magnesium ion also balances negative charge on the phosphate, making it an excellent leaving group.
In our examination of acyl substitution reactions, we will start with the formation and reactions of the highly reactive acyl phosphates. We will then discuss how thioesters play a key role in the acyl substitution reactions of lipid metabolism. Finally, we will take a look at some important acyl substitution reactions involving esters, as well as the formation and cleavage of the amide linkages in the peptide bonds of proteins. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/20%3A_Carboxylic_Acid_Derivatives/20.1%3A_Relative__Reactivities%2C_Structures%2C_and__Spectra__of_Carboxylic_Acid__Derivatives.txt |
Objectives
After completing this section, you should be able to
1. identify the reagent normally used to convert a carboxylic acid to an acid bromide.
2. write equations to show how an acid halide may be converted into each of the following: a carboxylic acid, an ester, an amide.
3. write a detailed mechanisms for the reaction of an acid halide with each of the following: water, an alcohol, ammonia, a primary or secondary amine.
4. identify the product formed when a given acid halide reacts with any of the following reagents: water, an alcohol, a primary or secondary amine.
5. identify the acid halide, the reagents, or both, needed to prepare a given carboxylic acid, ester or amide.
6. identify the product formed when a given acid halide reacts with water, a given alcohol, ammonia, or a given primary or secondary amine.
7. identify lithium aluminum hydride as a reagent for reducing acid halides to primary alcohols, and explain the limited practical value of this reaction.
8. identify the partial reduction of an acid halide using lithium tri‑tert‑butoxyaluminum to form an aldehyde.
9. write an equation to describe the formation of a tertiary alcohol by the reaction of an acid halide with a Grignard reagent.
10. write a detailed mechanism for the reaction of an acid halide with a Grignard reagent.
11. identify the product formed from the reaction of a given acid halide with a given Grignard reagent.
12. identify the acid halide, the Grignard reagent, or both, needed to prepare a given tertiary alcohol.
13. write an equation to illustrate the reaction of an acid halide with a lithium diorganocopper reagent.
14. identify the product formed from the reaction of a given acid halide with a given lithium diorganocopper reagent.
15. identify the acid halide, the lithium diorganocopper reagent, or both, that must be used to prepare a given ketone.
Study Notes
This figure provides a convenient general summary of a few of the reactions described in Section 21.4.
Note that LiAlH4 is a common reagent for hydride [H] reduction of and acid chloride to an alcohol.
Formation of Acid Halides
Carboxylic acids react with thionyl chloride (\(SOCl_2\)) to form acid chlorides. During the reaction the hydroxyl group of the carboxylic acid is converted to a chlorosulfite intermediate making it a better leaving group. The chloride anion produced during the reaction acts a nucleophile.
Earlier (Section 10.5) we saw that primary and secondary alcohols react with PBr3 to afford the corresponding alkyl bromide. In a similar fashion acid bromides can be formed from the carboxylic acid.
Nucleophilic Acyl Substitution Mechanism
If you understand the mechanism of a typical nucleophilic acyl substitution, the reaction of an acyl halide with water, an alcohol or ammonia should not present you with any difficulty.
X = Cl, Br :Nu = H2O, ROH, NH2R, NHR2 etc.
Mechanism
1) Nucleophilic attack by water
2) Leaving group is removed
3) Deprotonation
20.3: Chemistry of Carboxylic Anhydrides
Objectives
After completing this section, you should be able to
1. write an equation to illustrate the preparation of an acid anhydride from an acid halide and the sodium salt of a carboxylic acid.
2. identify the product formed from the reaction of a given acid halide with the sodium salt of a given carboxylic acid.
3. identify the acid halide, carboxylate salt, or both, required to prepare a given acid anhydride.
4. write an equation to describe the reaction of an acid anhydride with each of the following: water, alcohol, ammonia, a primary or secondary amine, lithium aluminum hydride.
5. identify the product formed when a given acid anhydride is reacted with any of the reagents listed in Objective 4, above.
6. write a detailed mechanism for the reaction of an acid anhydride with any of the reagents listed in Objective 4, above.
7. identify the acid anhydride, the nucleophilic reagent, or both, needed to prepare a specified carboxylic acid, ester, amide, or primary alcohol.
Study Notes
The reactions described in this section are, in principle, identical to those discussed in Section 21.4. Once you have understood the mechanism of nucleophilic acyl substitution, these reactions should not present you with any great difficulty, and memorization can be kept to a minimum.
This figure provides a convenient general summary of a few of the reactions described in Section 21.5. Note that from a synthetic perspective the ester‑ and amide‑forming reactions are the most common, so they are the focus of this section.
Preparation of Acid Anhydrides
Acid Chlorides react with carboxylic acids to form anhydrides
Acid Anhydrides react with alcohols to form esters
Reactions of anhydrides use Pyridine as a solvent
Mechanism
1) Nucleophilic Attack by the Alcohol
2) Deprotonation by pyridine
3) Leaving group removal
4) Protonation of the carboxylate
Mechanism
1) Nucleophilic Attack by the Amine
2) Deprotonation by the amine
3) Leaving group removal | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/20%3A_Carboxylic_Acid_Derivatives/20.2%3A_Chemistry_of_Alkanoyl__Halides.txt |
Objectives
After completing this section, you should be able to
1. discuss the wide occurrence of esters in nature, and their important commercial uses, giving one example of an ester linkage in nature, and one example of a commercially important ester.
2. write an equation to describe the hydrolysis of an ester under acidic or basic conditions.
3. identify the products formed from the hydrolysis of an given ester.
4. identify the reagents that can be used to bring about ester hydrolysis.
5. identify the structure of an unknown ester, given the products of its hydrolysis.
6. write the mechanism of alkaline ester hydrolysis.
7. write the mechanism of acidic ester hydrolysis.
8. write an equation to describe the reduction of an ester with lithium aluminum hydride.
9. identify the product formed from the reduction of a given ester (or lactone) with lithium aluminum hydride.
10. identify the ester, the reagents, or both, that should be used to prepare a given primary alcohol.
11. write a detailed mechanism for the reduction of an ester by lithium aluminum hydride.
12. identify diisobutylaluminum hydride as a reagent for reducing an ester to an aldehyde, and write an equation for such a reaction.
13. write an equation to describe the reaction of an ester with a Grignard reagent.
14. identify the product formed from the reaction of a given ester with a given Grignard reagent.
15. identify the ester, the Grignard reagent, or both, needed to prepare a given tertiary alcohol.
16. write a detailed mechanism for the reaction of an ester with a Grignard reagent.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• lactone
• saponification
Study Notes
Many esters have characteristic aromas and flavours. Some examples are listed below.
Basic structure:
Table 21.1 Structures and characteristic odours of selected esters
IUPAC name R R′ Aroma
octyl ethanoate CH3 CH3(CH2)6CH2 orange
propyl ethanoate CH3 CH3CH2 CH2 pear
2‑methylpropyl propanoate CH3CH2 (CH3)2CHCH2 rum
methyl butanoate CH3CH2CH2 CH3 apple
ethyl butanoate CH3CH2CH2 CH3CH2 pineapple
A “lactone” is a cyclic ester and has the general structure
By recognizing that the steps in the acidic hydrolysis of an ester are exactly the same as those in a Fischer esterification (but in the reverse order!), you can again minimize the amount of memorization that you must undertake. The details of both mechanisms can be deduced from the knowledge that both reactions are acid‑catalyzed nucleophilic acyl substitutions.
Introduction
Esters are readily synthesized and naturally abundant contributing to the flavors and aromas in many fruits and flowers.
They also make up the bulk of animal fats and vegetable oils—glycerides (fatty acid esters of glycerol). Soap is produced by a saponification (basic hydrolysis) reaction of a fat or oil.
Esters are also present in a number of important biological molecules and have several commercial and synthetic application. For example, polyester molecules make excellent fibers and are used in many fabrics. A knitted polyester tube, which is biologically inert, can be used in surgery to repair or replace diseased sections of blood vessels. PET is used to make bottles for soda pop and other beverages. It is also formed into films called Mylar. When magnetically coated, Mylar tape is used in audio- and videocassettes. Synthetic arteries can be made from PET, polytetrafluoroethylene, and other polymers.
The most important polyester, polyethylene terephthalate (PET), is made from terephthalic acid and ethylene glycol monomers:
Preparation of Esters
Carboxylic acids can react with alcohols to form esters
Acid chlorides react with alcohols to form esters
Acid Anhydrides react with alcohols to form esters
Conversion of Ester into Carboxylic acids: Hydrolysis
Esters can be cleaved back into a carboxylic acid and an alcohol by reaction with water and a catalytic amount of acid.
Mechanism
1) Protonation of the Carbonyl
2) Nucleophilic attack by water
3) Proton transfer
4) Leaving group removal
Esters can be cleaved back into a carboxylic acid and an alcohol by reaction with water and a base. The reaction is called a saponification from the Latin sapo which means soap. The name comes from the fact that soap used to me made by the ester hydrolysis of fats. Due to the basic conditions a carboxylate ion is made rather than a carboxylic acid.
Mechanism
1) Nucleophilic attack by hydroxide
2) Leaving group removal
3) Deprotonation
Conversion of Esters into Amides: Aminolysis
Esters reaction with ammonia and alkyl amines to yield amides.
Conversion of Ester into Alcohols: Reduction
Esters can be converted to 1o alcohols using LiAlH4, while sodium borohydride (\(NaBH_4\)) is not a strong enough reducing agent to perform this reaction.
Mechanism
1) Nucleophilic attack by the hydride
2) Leaving group removal
3) Nucleopilic attack by the hydride anion
4) The alkoxide is protonated
Esters can be converted to aldehydes using diisobutylaluminum hydride (DIBAH)
The reaction is usually carried out at -78 oC to prevent reaction with the aldehyde product.
Addition of Grignard reagents convert esters to 3o alcohols.
In effect the Grignard reagent adds twice.
Mechanism
1) Nucleophilic attack
2) Leaving group removal
3) Nucleophilic attack
4) Protonation
.
Contributors and Attributions
Charles Ophardt (Professor Emeritus, Elmhurst College); Virtual Chembook | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/20%3A_Carboxylic_Acid_Derivatives/20.4%3A_Chemistry_of_Esters.txt |
Objectives
After completing this section, you should be able to
1. write an equation to describe the preparation of an amide from an acid chloride.
2. identify the amide linkage as the basic unit from which all proteins are made, and hence recognize the importance of the amide linkage to biologists and biochemists.
3. write detailed mechanisms for the acidic and basic hydrolysis of amides.
4. write an equation to describe the reduction of an amide to an amine.
5. write a detailed mechanism for the reduction of an amide to an amine.
6. identify the product formed when a given amide is reduced with lithium aluminum hydride.
7. identify the amide, the reagents, or both, necessary to prepare a given amine by direct reduction.
8. identify lactams as being cyclic amides which undergo hydrolysis and reduction in a manner analogous to that of their acyclic counterparts.
Key Terms
Make certain that you can define, and use in context, the key term below.
• lactam
Study Notes
As the chapter which deals with amino acids and proteins is optional, it is possible for you to complete this course without studying these compounds in detail. However, because of their importance in biological systems, it is essential for all students completing this course to have some knowledge of their structure and properties.
When we talk about amino acids, we are generally referring to α‑amino acids; that is, compounds in which an amino (NH2) group and a carboxyl group are attached to the same carbon atom:
Notice that such compounds contain a chiral carbon atom (unless R = H). Proteins can be considered to consist of amino acid residues joined by amide (or peptide) links. These peptide links consist of exactly the same structural units that we find in secondary amides
Note: Although the terms secondary amide and tertiary amide have been used here, this usage is not in agreement with IUPAC recommendations. According to IUPAC, the secondary amide shown above should be referred to as an N‑substituted primary amide, and the tertiary amide should be referred to as a N, N‑disubstituted primary amide. IUPAC reserves the term secondary amide for compounds of the type
Because of the presence of the amide link in proteins,
we can expect the properties of these compounds to resemble those of secondary amides.
A lactam is a cyclic amide, in the same way that a lactone is a cyclic ester:
Among the most important molecules that contain lactam rings are the penicillins:
Preparation of Amides
Nitriles can be converted to amides. This reaction can be acid or base catalyzed
Carboxylic acid can be converted to amides by using DCC as an activating agent
Direct conversion of a carboxylic acid to an amide by reaction with an amine.
Acid chlorides react with ammonia, 1o amines and 2o amines to form amides
Acid Anhydrides react with ammonia, 1o amines and 2o amines to form amides
Conversion of Amides into Carboxylic Acids: Hydrolysis
This page describes the hydrolysis of amides under both acidic and alkaline conditions. It also describes the use of alkaline hydrolysis in testing for amides.
What is hydrolysis?
Technically, hydrolysis is a reaction with water. That is exactly what happens when amides are hydrolyzed in the presence of dilute acids such as dilute hydrochloric acid. The acid acts as a catalyst for the reaction between the amide and water. The alkaline hydrolysis of amides actually involves reaction with hydroxide ions, but the result is similar enough that it is still classed as hydrolysis.
Hydrolysis under acidic conditions
Taking ethanamide as a typical amide. If ethanamide is heated with a dilute acid (such as dilute hydrochloric acid), ethanoic acid is formed together with ammonium ions. So, if you were using hydrochloric acid, the final solution would contain ammonium chloride and ethanoic acid.
Hydrolysis under alkaline conditions
Also, if ethanamide is heated with sodium hydroxide solution, ammonia gas is given off and you are left with a solution containing sodium ethanoate.
Conversion of Amides into Amines: Reduction
Amides can be converted to 1°, 2° or 3° amines using LiAlH4.
Example 21.7.1: Amide Reductions
Alkyl groups attached to the nitrogen do not affect the reaction.
Mechanism
1) Nucleophilic attach by the hydride
2) Leaving group removal
3) Nucleophilic attach by the hydride
20.7: Amidates and Their Halogenation: The Hofmann Rearrangement
Hofmann rearrangement, also known as Hofmann degradation and not to be confused with Hofmann elimination, is the reaction of a primary amide with a halogen (chlorine or bromine) in strongly basic (sodium or potassium hydroxide) aqueous medium, which converts the amide to a primary amine. For example:
Mechanism:
20.8: Alkanenitriles: A Special Class of Carboxylic
Name the parent alkane (include the carbon atom of the nitrile as part of the parent) followed with the word -nitrile. The carbon in the nitrile is given the #1 location position. It is not necessary to include the location number in the name because it is assumed that the functional group will be on the end of the parent chain.
Cycloalkanes are followed by the word -carbonitrile. The substituent name is cyano.
• 1-butanenitrile or 1-cyanopropane
Try to name the following compounds using these conventions�
Try to draw structures for the following compounds�
• butanedinitrile J
• 2-methycyclohexanecarbonitrile J
Some common names that you should know are...
acetonitrile
benzonitrile
Try to draw a structure for the following compound�
• 2-methoxybenzonitrile J
Contributors
The electronic structure of nitriles is very similar to that of an alkyne with the main difference being the presence of a set of lone pair electrons on the nitrogen. Both the carbon and the nitrogen are sp hydridized which leaves them both with two p orbitals which overlap to form the two $\pi$ bond in the triple bond. The R-C-N bond angle in and nitrile is 180° which give a nitrile functional group a linear shape.
The lone pair electrons on the nitrogen are contained in a sp hybrid orbital which makes them much less basic and an amine. The 50% character of an sp hybrid orbital close to the nucleus and therefore less basic compared to other nitrogen containing compounds such as amines.
The presence of an electronegative nitrogen causes nitriles to be very polar molecules. Consequently, nitriles tend to have higher boiling points than molecules with a similar size.
Contributors
Nitriles can be converted to carboxylic acid with heating in sulfuric acid. During the reaction an amide intermediate is formed.
Contributors
Prof. Steven Farmer (Sonoma State University)
Grignard reagents can attack the electophillic carbon in a nitrile to form an imine salt. This salt can then be hydrolyzed to become a ketone.
Mechanism
1) Nucleophilic Attack by the Grignard Reagent
2) Protonation
3) Protonation
4) Nucleophilic attack by water
5) Proton Transfer
6) Leaving group removal
7) Deprotonation
Contributors
Prof. Steven Farmer (Sonoma State University)
Nitriles can be converted to 1° amines by reaction with LiAlH4. During this reaction the hydride nucleophile attacks the electrophilic carbon in the nitrile to form an imine anion. Once stabilized by a Lewis acid-base complexation the imine salt can accept a second hydride to form a dianion. The dianion can then be converted to an amine by addition of water.
Mechanism
1) Nucleophilic Attack by the Hydride
2) Second nucleophilic attack by the hydride.
3) Protonation by addition of water to give an amine
Contributors
Prof. Steven Farmer (Sonoma State University) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/20%3A_Carboxylic_Acid_Derivatives/20.6%3A_Amides%3A__The__Least__Reactive__Carboxylic_Acid__Derivatives.txt |
Objectives
After completing this section, you should be able to
1. classify a given amine as being primary, secondary or tertiary.
2. explain, briefly, the difference in meaning of the terms primary, secondary and tertiary when they are applied to the structures of amines and alcohols.
3. determine whether a given structure represents a quaternary ammonium cation.
4. provide an acceptable IUPAC name for an amine, given its Kekulé, condensed or shorthand structure.
5. draw the structure of an amine, given its IUPAC name.
6. give the name and structure of one typical heterocyclic amine (e.g., pyridine).
Key Terms
Make certain that you can define, and use in context, the key terms below.
• amine
• primary amine
• secondary amine
• quaternary ammonium cation
• tertiary amine
Study Notes
You should recognize that heterocyclic amines—compounds in which the nitrogen atom occurs as part of a ring—are very common in organic chemistry. Be prepared to meet such compounds throughout this, and subsequent chapters, but do not try to memorize all of the names and structures givenin the reading. You should, however, commit the structures of pyridine and pyrrole to memory.
Amines are derivatives of ammonia in which one or more of the hydrogens has been replaced by an alkyl or aryl group. The nomenclature of amines is complicated by the fact that several different nomenclature systems exist, and there is no clear preference for one over the others. Furthermore, the terms primary (1º), secondary (2º) & tertiary (3º) are used to classify amines in a completely different manner than they were used for alcohols or alkyl halides. When applied to amines these terms refer to the number of alkyl (or aryl) substituents bonded to the nitrogen atom, whereas in other cases they refer to the nature of an alkyl group. The four compounds shown in the top row of the following diagram are all C4H11N isomers. The first two are classified as 1º-amines, since only one alkyl group is bonded to the nitrogen; however, the alkyl group is primary in the first example and tertiary in the second. The third and fourth compounds in the row are 2º and 3º-amines respectively. A nitrogen bonded to four alkyl groups will necessarily be positively charged, and is called a quaternary (4º)-ammonium cation. For example, (CH3)4N(+) Br(–) is tetramethylammonium bromide.
• The IUPAC names are listed first and colored blue. This system names amine functions as substituents on the largest alkyl group. The simple -NH substituent found in 1º-amines is called an amino group. For 2º and 3º-amines a compound prefix (e.g. dimethylamino in the fourth example) includes the names of all but the root alkyl group.
• The Chemical Abstract Service has adopted a nomenclature system in which the suffix -amine is attached to the root alkyl name. For 1º-amines such as butanamine (first example) this is analogous to IUPAC alcohol nomenclature (-ol suffix). The additional nitrogen substituents in 2º and 3º-amines are designated by the prefix N- before the group name. These CA names are colored magenta in the diagram.
• Finally, a common system for simple amines names each alkyl substituent on nitrogen in alphabetical order, followed by the suffix -amine. These are the names given in the last row (colored black).
Many aromatic and heterocyclic amines are known by unique common names, the origins of which are often unknown to the chemists that use them frequently. Since these names are not based on a rational system, it is necessary to memorize them. There is a systematic nomenclature of heterocyclic compounds, but it will not be discussed here. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/21%3A_Amines_and_Their_Derivatives/21.01%3A_Naming__the_Amines.txt |
Objectives
After completing this section, you should be able to
1. describe the geometry and bonding of simple amines.
2. explain why most chiral amines cannot be resolved into their two enantiomers.
Key Terms
Make certain that you can define, and use in context, the key term below.
• pyramidal inversion
Study Notes
Molecular models may help you to understand pyramidal inversion.
Amines typically have three bonds and one pair of lone pair electrons. This makes the nitrogen sp3 hybridized, trigonal pyramidal, with a bond angle of roughly 109.5o.
Stereogenic Nitrogen
Single-bonded nitrogen is pyramidal in shape, with the non-bonding electron pair pointing to the unoccupied corner of a tetrahedral region. Since the nitrogen in these compounds is bonded to three different groups, its configuration is chiral. The non-identical mirror-image configurations are illustrated in the following diagram (the remainder of the molecule is represented by R, and the electron pair is colored yellow). If these configurations were stable, there would be four additional stereoisomers of ephedrine and pseudoephedrine. However, pyramidal nitrogen is normally not configurationally stable. It rapidly inverts its configuration (equilibrium arrows) by passing through a planar, sp2-hybridized transition state, leading to a mixture of interconverting R and S configurations. If the nitrogen atom were the only chiral center in the molecule, a 50:50 (racemic) mixture of R and S configurations would exist at equilibrium. If other chiral centers are present, as in the ephedrin isomers, a mixture of diastereomers will result. The take-home message is that nitrogen does not contribute to isolable stereoisomers.
Boiling Point and Water Solubility
It is instructive to compare the boiling points and water solubility of amines with those of corresponding alcohols and ethers. The dominant factor here is hydrogen bonding, and the first table below documents the powerful intermolecular attraction that results from -O-H---O- hydrogen bonding in alcohols (light blue columns). Corresponding -N-H---N- hydrogen bonding is weaker, as the lower boiling points of similarly sized amines (light green columns) demonstrate. Alkanes provide reference compounds in which hydrogen bonding is not possible, and the increase in boiling point for equivalent 1º-amines is roughly half the increase observed for equivalent alcohols.
Compound CH3CH3 CH3OH CH3NH2 CH3CH2CH3 CH3CH2OH CH3CH2NH2
Mol.Wt. 30 32 31 44 46 45
Boiling
Point ºC
-88.6º 65º -6.0º -42º 78.5º 16.6º
The second table illustrates differences associated with isomeric 1º, 2º & 3º-amines, as well as the influence of chain branching. Since 1º-amines have two hydrogens available for hydrogen bonding, we expect them to have higher boiling points than isomeric 2º-amines, which in turn should boil higher than isomeric 3º-amines (no hydrogen bonding). Indeed, 3º-amines have boiling points similar to equivalent sized ethers; and in all but the smallest compounds, corresponding ethers, 3º-amines and alkanes have similar boiling points. In the examples shown here, it is further demonstrated that chain branching reduces boiling points by 10 to 15 ºC.
Compound CH3(CH2)2CH3 CH3(CH2)2OH CH3(CH2)2NH2 CH3CH2NHCH3 (CH3)3CH (CH3)2CHOH (CH3)2CHNH2 (CH3)3N
Mol.Wt. 58 60 59 59 58 60 59 59
Boiling
Point ºC
-0.5º 97º 48º 37º -12º 82º 34º
The water solubility of 1º and 2º-amines is similar to that of comparable alcohols. As expected, the water solubility of 3º-amines and ethers is also similar. These comparisons, however, are valid only for pure compounds in neutral water. The basicity of amines (next section) allows them to be dissolved in dilute mineral acid solutions, and this property facilitates their separation from neutral compounds such as alcohols and hydrocarbons by partitioning between the phases of non-miscible solvents. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/21%3A_Amines_and_Their_Derivatives/21.02%3A_Structural_and__Physical__Properties_of_Amines.txt |
Objectives
After completing this section, you should be able to
1. identify the region of the infrared spectrum that shows absorptions resulting from the N$\ce{-}$H bonds of primary and secondary amines.
2. describe a characteristic change that occurs in the infrared spectrum of an amine when a small amount of mineral acid is added to the sample.
3. use 1H NMR spectra in determining the structure of an unknown amine.
4. use the “nitrogen rule” of mass spectrometry to determine whether a compound has an odd or even number of nitrogen atoms in its structure.
5. predict the prominent peaks in the mass spectrum of a given amine.
6. use the mass spectrum of an unknown amine in determining its structure.
Key Terms
Make certain that you can define, and use in context, the key term below.
• nitrogen rule
Study Notes
You should note the spectroscopic similarities between amines and alcohols: both have infrared absorptions in the 3300–3360 cm−1 region, and in both cases, the proton that is attached to the heteroatom gives rise to an often indistinct signal in the 1H NMR spectrum.
NMR
The hydrogens attached to an amine show up ~ 0.5-5.0 ppm. The location is dependent on the amount of hydrogen bonding and the sample's concentration.
The hydrogens on carbons directly bonded to an amine typically appear ~2.3-3.0 ppm.
Addition of D2O will normally cause all hydrogens on non-carbon atoms to exchange with deuteriums, thus making these resonances "disappear." Addition of a few drops of D2O causing a signal to vanish can help confirm the presence of -NH.
IR
The infrared spectrum of aniline is shown beneath the following table. Some of the characteristic absorptions for C-H stretching and aromatic ring substitution are also marked, but not colored.
Amine Class
Stretching Vibrations
Bending Vibrations
Primary (1°)
The N-H stretching absorption is less sensitive to hydrogen bonding than are O-H absorptions. In the gas phase and in dilute CCl4 solution free N-H absorption is observed in the 3400 to 3500 cm-1 region. Primary aliphatic amines display two well-defined peaks due to asymmetric (higher frequency) and symmetric N-H stretching, separated by 80 to 100 cm-1. In aromatic amines these absorptions are usually 40 to 70 cm-1 higher in frequency. A smaller absorption near 3200 cm-1 (shaded orange in the spectra) is considered to be the result of interaction between an overtone of the 1600 cm-1 band with the symmetric N-H stretching band.
C-N stretching absorptions are found at 1200 to 1350 cm-1 for aromatic amines, and at 1000 to 1250 cm-1 for aliphatic amines.
Strong in-plane NH2 scissoring absorptions at 1550 to 1650 cm-1, and out-of-plane wagging at 650 to 900 cm-1 (usually broad) are characteristic of 1°-amines.
Secondary (2°)
Secondary amines exhibit only one absorption near 3420 cm-1. Hydrogen bonding in concentrated liquids shifts these absorptions to lower frequencies by about 100 cm-1. Again, this absorption appears at slightly higher frequency when the nitrogen atom is bonded to an aromatic ring.
The C-N absorptions are found in the same range, 1200 to 1350 cm-1(aromatic) and 1000 to 1250 cm-1 (aliphatic) as for 1°-amines.
A weak N-H bending absorption is sometimes visible at 1500 to 1600 cm-1. A broad wagging absorption at 650 to 900 cm-1 may be discerned in liquid film samples.
Tertiary (3°)
No N-H absorptions. The C-N absorptions are found in the same range, 1200 to 1350 cm-1 (aromatic) and 1000 to 1250 cm-1 (aliphatic) as for 1°-amines.
Aside from the C-N stretch noted on the left, these compounds have spectra characteristic of their alkyl and aryl substituents.
Nitrogen Rule
The nitrogen rule states that a molecule that has no or even number of nitrogen atoms has an even nominal mass, whereas a molecule that has an odd number of nitrogen atoms has an odd nominal mass.
eg. 1:
eg. 2:
eg. 3: | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/21%3A_Amines_and_Their_Derivatives/21.03%3A_Spectroscopy_of__the_Amine__Group.txt |
Objectives
After completing this section, you should be able to
1. account for the basicity and nucleophilicity of amines.
2. explain why amines are more basic than amides, and better nucleophiles.
3. describe how an amine can be extracted from a mixture that also contains neutral compounds illustrating the reactions which take place with appropriate equations.
4. explain why primary and secondary (but not tertiary) amines may be regarded as very weak acids, and illustrate the synthetic usefulness of the strong bases that can be formed from these weak acids.
Key Terms
Make certain that you can define, and use in context, the key term below.
• amide
Study Notes
The lone pair of electrons on the nitrogen atom of amines makes these compounds not only basic, but also good nucleophiles. Indeed, we have seen in past chapters that amines react with electrophiles in several polar reactions (see for example the nucleophilic addition of amines in the formation of imines and enamines in Section 19.8).
The ammonium ions of most simple aliphatic amines have a pKa of about 10 or 11. However, these simple amines are all more basic (i.e., have a higher pKa) than ammonia. Why? Remember that, relative to hydrogen, alkyl groups are electron releasing, and that the presence of an electron‑releasing group stabilizes ions carrying a positive charge. Thus, the free energy difference between an alkylamine and an alkylammonium ion is less than the free energy difference between ammonia and an ammonium ion; consequently, an alkylamine is more easily protonated than ammonia, and therefore the former has a higher pKa than the latter.
Basicity of nitrogen groups
In this section we consider the relative basicity of amines. When evaluating the basicity of a nitrogen-containing organic functional group, the central question we need to ask ourselves is: how reactive (and thus how basic and nucleophilic) is the lone pair on the nitrogen? In other words, how much does that lone pair want to break away from the nitrogen nucleus and form a new bond with a hydrogen. The lone pair electrons makes the nitrogen in amines electron dense, which is represents by a red color in the electrostatic potential map present below left. Amine are basic and easily react with the hydrogen of acids which are electron poor as seen below.
Amines are one of the only neutral functional groups which are considered bases which is a consequence of the presence of the lone pair electrons on the nitrogen. During an acid/base reaction the lone pair electrons attack an acidic hydrogen to form a N-H bond. This gives the nitrogen in the resulting ammonium salt four single bonds and a positive charge.
Amines react with water to establish an equilibrium where a proton is transferred to the amine to produce an ammonium salt and the hydroxide ion, as shown in the following general equation:
$RNH2_{(aq)}+H_2O_{(l)} \rightleftharpoons RNH3^+_{(aq)}+OH^−_{(aq)} \label{16.5.4}$
The equilibrium constant for this reaction is the base ionization constant (Kb), also called the base dissociation constant:
$K_b=\dfrac{[RNH3^+][OH^−]}{[NH2]} \label{16.5.5}$
pKb = -log Kb
Just as the acid strength of a carboxylic acid can be measured by defining an acidity constant Ka (Section 2-8), the base strength of an amine can be measured by defining an analogous basicity constant Kb. The larger the value of Kb and the smaller the value of pKb, the more favorable the proton-transfer equilibrium and the stronger the base.
However, Kb values are often not used to discuss relative basicity of amines. It is common to compare basicity's of amines by using the Ka's of their conjugate acids, which is the corresponding ammonium ion. Fortunately, the Ka and Kb values for amines are directly related.
Consider the reactions for a conjugate acid-base pair, RNH3+ − RNH2:
$\ce{RNH3+}(aq)+\ce{H2O}(l)⇌\ce{RNH2}(aq)+\ce{H3O+}(aq) \hspace{20px} K_\ce{a}=\ce{\dfrac{[RNH2][H3O]}{[RNH3+]}}$
$\ce{RNH2}(aq)+\ce{H2O}(l)⇌\ce{RNH3+}(aq)+\ce{OH-}(aq) \hspace{20px} K_\ce{b}=\ce{\dfrac{[RNH3+][OH-]}{[RNH2]}}$
Adding these two chemical equations together yields the equation for the autoionization for water:
$\cancel{\ce{RNH3+}(aq)}+\ce{H2O}(l)+\cancel{\ce{RNH2}(aq)}+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\cancel{\ce{RNH2}(aq)}+\ce{OH-}(aq)+\cancel{\ce{RNH3+}(aq)}$
$\ce{2H2O}(l)⇌\ce{H3O+}(aq)+\ce{OH-}(aq)$
Given that the K expression for a chemical equation formed from adding two or more other equations is the mathematical product of the input equations’ K constants.
Ka X Kb = {2 H2O} / (H3O+}{OH-} = Kw
$K_\ce{a}=\dfrac{K_\ce{w}}{K_\ce{b}}$
pKa + pKb =14
Thus if the Ka for an ammonium ion is know the Kb for the corresponding amine can be calculated using the equation Kb = Kw / Ka. This relationship shows that as an ammonium ion becomes more acidic (Ka increases / pKa decreases) the correspond base becomes weaker (Kb decreases / pKb increases)
Weaker Base = Larger Ka and Smaller pKa of the Ammonium ion
Stronger Base = Smaller Ka and Larger pKa of the Ammonium ion
Like ammonia, most amines are Brønsted-Lowry and Lewis bases, but their base strength can be changed enormously by substituents. Most simple alkyl amines have pKa's in the range 9.5 to 11.0, and their aqueous solutions are basic (have a pH of 11 to 12, depending on concentration).
Aromatic herterocyclic amines (such as pyrimidine, pyridine, imidazole, pyrrole) are significantly weaker bases as a consequence of three factors. The first of these is the hybridization of the nitrogen. In each case the heterocyclic nitrogen is sp2 hybridized. The increasing s-character brings it closer to the nitrogen nucleus, reducing its tendency to bond to a proton compared to sp3 hybridized nitrogens. The very low basicity of pyrrole reflects the exceptional delocalization of the nitrogen electron pair associated with its incorporation in an aromatic ring. Imidazole (pKa = 6.95) is over a million times more basic than pyrrole because the sp2 nitrogen that is part of one double bond is structurally similar to pyridine, and has a comparable basicity.
Basicity of common amines (pKa of the conjugate ammonium ions)
Inductive Effects in Nitrogen Basicity
Alkyl groups donate electrons to the more electronegative nitrogen. The inductive effect makes the electron density on the alkylamine's nitrogen greater than the nitrogen of ammonia. The small amount of extra negative charge built up on the nitrogen atom makes the lone pair even more attractive towards hydrogen ions. Correspondingly, primary, secondary, and tertiary alkyl amines are more basic than ammonia.
Compound pKa of the conjugate acid
NH3 9.3
CH3NH2 10.66
(CH3)2NH 10.74
(CH3)3N 9.81
Comparing the Basicity of Alkylamines to Amides
The nitrogen atom is strongly basic when it is in an amine, but not significantly basic when it is part of an amide group. While the electron lone pair of an amine nitrogen is localized in one place, the lone pair on an amide nitrogen is delocalized by resonance. The electron density – in the form of a lone pair – is stabilized by resonance delocalization, even though there is not a negative charge involved. Here’s another way to think about it: the lone pair on an amide nitrogen is not as available for bonding with a proton – these two electrons are too stable being part of the delocalized pi-bonding system. The electrostatic potential map shows the effect of resonance on the basicity of an amide. The map shows that the electron density, shown in red, is almost completely shifted towards the oxygen. This greatly decreases the basicity of the lone pair electrons on the nitrogen in an amide.
Comparison of amines and amides to rationalize the the pKa values of their conjugate acids
Amine Extraction in the Laboratory
Extraction is often employed in organic chemistry to purify compounds. Liquid-liquid extractions take advantage of the difference in solubility of a substance in two immiscible liquids (e.g. ether and water). The two immiscible liquids used in an extraction process are (1) the solvent in which the solids are dissolved, and (2) the extracting solvent. The two immiscible liquids are then easily separated using a separatory funnel. For amines one can take advantage of their basicity by forming the protonated salt (RNH2+Cl), which is soluble in water. The salt will extract into the aqueous phase leaving behind neutral compounds in the non-aqueous phase. The aqueous layer is then treated with a base (NaOH) to regenerate the amine and NaCl. A second extraction-separation is then done to isolate the amine in the non-aqueous layer and leave behind NaCl in the aqueous layer.
Important Reagent Bases
The significance of all these acid-base relationships to practical organic chemistry lies in the need for organic bases of varying strength, as reagents tailored to the requirements of specific reactions. The common base sodium hydroxide is not soluble in many organic solvents, and is therefore not widely used as a reagent in organic reactions. Most base reagents are alkoxide salts, amines or amide salts. Since alcohols are much stronger acids than amines, their conjugate bases are weaker than amine bases, and fill the gap in base strength between amines and amide salts.
Base Name Pyridine Triethyl
Amine
Hünig's Base Barton's
Base
Potassium
t-Butoxide
Sodium HMDS LDA
Formula (C2H5)3N (CH3)3CO(–) K(+) [(CH3)3Si]2N(–) Na(+) [(CH3)2CH]2N(–) Li(+)
pKa of conjugate acid 5.3 10.7 11.4 14 19 26 35.7
Basicity of common amines (pKa of the conjugate ammonium ions)
Pyridine is commonly used as an acid scavenger in reactions that produce mineral acid co-products. Its basicity and nucleophilicity may be modified by steric hindrance, as in the case of 2,6-dimethylpyridine (pKa=6.7), or resonance stabilization, as in the case of 4-dimethylaminopyridine (pKa=9.7). Hünig's base is relatively non-nucleophilic (due to steric hindrance), and is often used as the base in E2 elimination reactions conducted in non-polar solvents. Barton's base is a strong, poorly-nucleophilic, neutral base that serves in cases where electrophilic substitution of other amine bases is a problem. The alkoxides are stronger bases that are often used in the corresponding alcohol as solvent, or for greater reactivity in DMSO. Finally, the two amide bases see widespread use in generating enolate bases from carbonyl compounds and other weak carbon acids.
In addition to acting as a base, 1o and 2o amines can act as very weak acids. Their N-H proton can be removed if they are reacted with a strong enough base. An example is the formation of lithium diisopropylamide (LDA, LiN[CH(CH3)2]2) by reacting n-butyllithium with diisopropylamine (pKa 36) (Section 22-5). LDA is a very strong base and is commonly used to create enolate ions by deprotonating an alpha-hydrogen from carbonyl compounds (Section 22-7).
Contributors and Attributions
Objectives
After completing this section, you should be able to
1. use the concept of resonance to explain why arylamines are less basic than their aliphatic counterparts.
2. arrange a given series of arylamines in order of increasing or decreasing basicity.
3. discuss, in terms of inductive and resonance effects, why a given arylamine is more or less basic than aniline.
Study Notes
With reference to the discussion of base strength, the traditional explanation for the base‑strengthening effect of electron‑releasing (I) substituents is that such substituents help to stabilize the positive charge on an arylammonium ion more than they stabilize the unprotonated compound, thereby lowering ΔG°.
The electron‑withdrawing (i.e., deactivating) substituents decrease the stability of a positively charged arylammonium ion.
Note that the arylammonium ion derived from aniline, PhNH3+, is commonly referred to as the anilinium ion.
Basicity of Aniline
Aniline is substantially less basic than methylamine, as is evident by looking at the pKa values for their respective ammonium conjugate acids (remember that the lower the pKa of the conjugate acid, the weaker the base).
This difference is basicity can be explained by the observation that, in aniline, the lone pair of electrons on the nitrogen are delocalized by the aromatic p system, making it less available for bonding to H+ and thus less basic. The lone pair electrons of aniline are involved in four resonance forms making them more stable and therefore less reactive relative to alkylamines.
The effect of delocalization can be seen when viewing the electrostatic potential maps of aniline an methyl amine. The nitrogen of methyl amine has a significant amount of electron density on its nitrogen, shown as a red color, which accounts for it basicity compared to aniline. While the electron density of aniline's nitrogen is delocalized in the aromatic ring making it less basic.
Basicity of Substituted Arylamines
The addition of substituents onto the aromatic ring can can make arylamines more or less basic. Substituents which are electron-withdrawing (-Cl, -CF3, -CN, -NO2) decrease the electron density in the aromatic ring and on the amine making the arylamine less basic. In particular, the nitro group of para-nitroaniline allows for an additional resonance form to be drawn, which further stabilizes the lone pair electrons from the nitrogen, making the substituted arylamine less basic than aniline. This effect is analogous to the one discussed for the acidity of substituted phenols in Section 17.2.
Substituents which are electron-donating (-CH3, -OCH3, -NH2) increase the electron density in the aromatic ring and on the amine making the arylamine more basic. In the case of para-methoxyaniline, the lone pair on the methoxy group donates electron density to the aromatic system, and a resonance contributor can be drawn in which a negative charge is placed on the carbon adjacent to the nitrogen, which makes the substituted arylamine more basic than aniline.
Increased Basicity of para-Methoxyaniline due to Electron-Donation
The shifting electron density of aniline, p-nitroaniline, and p-methoxyaniline are seen in their relative electrostatic potential maps. For p-Nitroaniline virtually all of the electron density, shown as a red/yellow color. is pulled toward the electron-withdrawing nitro group. In p-methoxyaninline the electron donating methoxy group donates electron density into the ring. The amine in p-methoxyaniline is shown to have more electron density, shown as a yellow color, when compared to the amine in aniline.
Exercise s$1$
1) Using the knowledge of the electron donating or withdrawing effects of subsituents gained in Section 16.6, rank the following compound in order of decreasing basicity.
a) p-Nitroaniline, methyl p-aminobenzoate, p-chloroaniline
b) p-Bromoaniline, p-Aminobenzonitrile, p-ethylaniline
c) p-(Trifluoromethyl)aniline, p-methoxyaniline, p-methylaniline
Answers
1)
a) p-Chloroaniline, methyl p-aminobenzoate, p-nitroaniline
b) p-Ethylaniline, p-Bromoaniline, p-aminobenzonitrile
c) p-Methoxyaniline, p-methylaniline, p-(trifluoromethyl)aniline
Acidity of Amines
We normally think of amines as bases, but it must be remembered that 1º and 2º-amines are also very weak acids (ammonia has a pKa = 34). In this respect it should be noted that pKa is being used as a measure of the acidity of the amine itself rather than its conjugate acid, as in the previous section. For ammonia this is expressed by the following hypothetical equation:
NH3 + H2O ____> NH2(–) + H2O-H(+)
The same factors that decreased the basicity of amines increase their acidity. This is illustrated by the following examples, which are shown in order of increasing acidity. It should be noted that the first four examples have the same order and degree of increased acidity as they exhibited decreased basicity in the previous table. The first compound is a typical 2º-amine, and the three next to it are characterized by varying degrees of nitrogen electron pair delocalization. The last two compounds (shaded blue) show the influence of adjacent sulfonyl and carbonyl groups on N-H acidity. From previous discussion it should be clear that the basicity of these nitrogens is correspondingly reduced.
Compound C6H5SO2NH2
pKa 33 27 19 15 10 9.6
The acids shown here may be converted to their conjugate bases by reaction with bases derived from weaker acids (stronger bases). Three examples of such reactions are shown below, with the acidic hydrogen colored red in each case. For complete conversion to the conjugate base, as shown, a reagent base roughly a million times stronger is required.
C6H5SO2NH2 + KOH C6H5SO2NH(–) K(+) + H2O a sulfonamide base
(CH3)3COH + NaH (CH3)3CO(–) Na(+) + H2 an alkoxide base
(C2H5)2NH + C4H9Li (C2H5)2N(–) Li(+) + C4H10 an amide base | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/21%3A_Amines_and_Their_Derivatives/21.04%3A_Acidity__and__Basicity__of_Amines.txt |
This page summarises the reactions of amines as nucleophiles. This includes their reactions with halogenoalkanes (haloalkanes or alkyl halides), with acyl chlorides (acid chlorides) and with acid anhydrides.
Amines by direct nucleophilic substitution
A nucleophile is something which is attracted to, and then attacks, a positive or slightly positive part of another molecule or ion. All amines contain an active lone pair of electrons on the very electronegative nitrogen atom. It is these electrons which are attracted to positive parts of other molecules or ions.
The reactions of primary amines with halogenoalkanes
You get a complicated series of reactions on heating to give a mixture of products - probably one of the most confusing sets of reactions you will meet at this level. The products of the reactions include secondary and tertiary amines and their salts, and quaternary ammonium salts.
Making secondary amines and their salts
In the first stage of the reaction, you get the salt of a secondary amine formed. For example if you started with ethylamine and bromoethane, you would get diethylammonium bromide
In the presence of excess ethylamine in the mixture, there is the possibility of a reversible reaction. The ethylamine removes a hydrogen from the diethylammonium ion to give free diethylamine - a secondary amine.
Making tertiary amines and their salts
But it doesn't stop here! The diethylamine also reacts with bromoethane - in the same two stages as before. This is where the reaction would start if you reacted a secondary amine with a halogenoalkane.
In the first stage, you get triethylammonium bromide.
There is again the possibility of a reversible reaction between this salt and excess ethylamine in the mixture.
The ethylamine removes a hydrogen ion from the triethylammonium ion to leave a tertiary amine - triethylamine.
Making a quaternary ammonium salt
The final stage! The triethylamine reacts with bromoethane to give tetraethylammonium bromide - a quaternary ammonium salt (one in which all four hydrogens have been replaced by alkyl groups).
This time there isn't any hydrogen left on the nitrogen to be removed. The reaction stops here.
Preparation of Primary Amines
Although direct alkylation of ammonia by alkyl halides leads to 1º-amines, alternative procedures are preferred in many cases. These methods require two steps, but they provide pure product, usually in good yield. The general strategy is to first form a carbon-nitrogen bond by reacting a nitrogen nucleophile with a carbon electrophile. The following table lists several general examples of this strategy in the rough order of decreasing nucleophilicity of the nitrogen reagent. In the second step, extraneous nitrogen substituents that may have facilitated this bonding are removed to give the amine product.
Nitrogen
Reactant
Carbon
Reactant
1st Reaction
Type
Initial Product
2nd Reaction
Conditions
2nd Reaction
Type
Final Product
N3(–) RCH2-X or
R2CH-X
SN2 RCH2-N3 or
R2CH-N3
LiAlH4 or
4 H2 & Pd
Hydrogenolysis RCH2-NH2 or
R2CH-NH2
C6H5SO2NH(–) RCH2-X or
R2CH-X
SN2 RCH2-NHSO2C6H5 or
R2CH-NHSO2C6H5
Na in NH3 (liq) Hydrogenolysis RCH2-NH2 or
R2CH-NH2
CN(–) RCH2-X or
R2CH-X
SN2 RCH2-CN or
R2CH-CN
LiAlH4 Reduction RCH2-CH2NH2 or
R2CH-CH2NH2
NH3 RCH=O or
R2C=O
Addition /
Elimination
RCH=NH or
R2C=NH
H2 & Ni
or NaBH3CN
Reduction RCH2-NH2 or
R2CH-NH2
NH3 RCOX Addition /
Elimination
RCO-NH2 LiAlH4 Reduction RCH2-NH2
NH2CONH2
(urea)
R3C(+) SN1 R3C-NHCONH2 NaOH soln. Hydrolysis R3C-NH2
A specific example of each general class is provided in the diagram below. In the first two, an anionic nitrogen species undergoes an SN2 reaction with a modestly electrophilic alkyl halide reactant. For example #2 an acidic phthalimide derivative of ammonia has been substituted for the sulfonamide analog listed in the table. The principle is the same for the two cases, as will be noted later. Example #3 is similar in nature, but extends the carbon system by a methylene group (CH2). In all three of these methods 3º-alkyl halides cannot be used because the major reaction path is an E2 elimination.
The methods illustrated by examples #4 and #5 proceed by attack of ammonia, or equivalent nitrogen nucleophiles, at the electrophilic carbon of a carbonyl group. A full discussion of carbonyl chemistry is presented later, but for present purposes it is sufficient to recognize that the C=O double bond is polarized so that the carbon atom is electrophilic. Nucleophile addition to aldehydes and ketones is often catalyzed by acids. Acid halides and anhydrides are even more electrophilic, and do not normally require catalysts to react with nucleophiles. The reaction of ammonia with aldehydes or ketones occurs by a reversible addition-elimination pathway to give imines (compounds having a C=N function). These intermediates are not usually isolated, but are reduced as they are formed (i.e. in situ). Acid chlorides react with ammonia to give amides, also by an addition-elimination path, and these are reduced to amines by LiAlH4.
The 6th example is a specialized procedure for bonding an amino group to a 3º-alkyl group (none of the previous methods accomplishes this). Since a carbocation is the electrophilic species, rather poorly nucleophilic nitrogen reactants can be used. Urea, the diamide of carbonic acid, fits this requirement nicely. The resulting 3º-alkyl-substituted urea is then hydrolyzed to give the amine.
One important method of preparing 1º-amines, especially aryl amines, uses a reverse strategy. Here a strongly electrophilic nitrogen species (NO2(+)) bonds to a nucleophilic carbon compound. This nitration reaction gives a nitro group that can be reduced to a 1º-amine by any of several reduction procedures.
The Hofmann rearrangement of 1º-amides provides an additional synthesis of 1º-amines. To learn about this useful procedure Click Here.
Reduction of Nitro Groups
Several methods for reducing nitro groups to amines are known. These include catalytic hydrogenation (H2 + catalyst), zinc or tin in dilute mineral acid, and sodium sulfide in ammonium hydroxide solution. The procedures described above are sufficient for most case
Nitriles can be converted to 1° amines by reaction with LiAlH4
During this reaction the hydride nucleophile attacks the electrophilic carbon in the nitrile to form an imine anion. Once stabilized by a Lewis acid-base complexation the imine salt can accept a second hydride to form a dianion. The dianion can then be converted to an amine by addition of water.
Reductive amination
Aldehydes and ketones can be converted into 1o, 2o and 3o amines using reductive amination. The reaction takes place in two parts. The first step is the nucleophiic addition of the carbonyl group to form an imine. The second step is the reduction of the imine to an amine using an reducing agent. A reducing agent commonly used for this reaction is sodium cyanoborohydride (NaBH3CN).
21.06: Synthesis of Amines by Reductive Amination
Reductive Amination
Aldehydes and ketones can be converted into 1o, 2o and 3o amines using reductive amination. The reaction takes place in two parts. The first step is the nucleophiic addition of the carbonyl group to form an imine. The second step is the reduction of the imine to an amine using an reducing agent. A reducing agent commonly used for this reaction is sodium cyanoborohydride (NaBH3CN). | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/21%3A_Amines_and_Their_Derivatives/21.05%3A_Synthesis_of_Amines__by_Alkylation.txt |
Objectives
After completing this section, you should be able to
1. write equations to illustrate the synthesis of amines by
1. reduction of nitriles or amides and nitro compounds.
2. reactions involving alkyl groups:
1. SN2 reactions of alkyl halides, ammonia and other amines.
2. nucleophilic attack by an azide ion on an alkyl halide, followed by reduction of the azide so formed.
3. alkylation of potassium phthalimide, followed by hydrolysis of the N‑alkyl phthalimide so formed (i.e., the Gabriel synthesis).
3. reductive amination of aldehydes or ketones.
4. Hofmann or Curtius rearrangements.
2. write detailed mechanisms for each of the steps involved in the synthetic routes outlined in Objective 1.
3. identify the product or products formed when
1. a given nitrile or amide is reduced using lithium aluminum hydride.
2. a given alkyl halide is reacted with ammonia or an alkylamine.
3. a given alkyl halide is reacted with azide ion and the resulting product is reduced.
4. a given alkyl halide is reacted with potassium phthalimide and the resulting product is hydrolyzed.
5. a given aldehyde or ketone is reacted with ammonia or an amine in the presence of nickel catalyst.
6. a given amide is treated with halogen and base.
7. a given acyl azide is heated and then hydrolyzed.
4. identify the starting material, the other reagents, or both, needed to synthesize a given amine by any of the routes listed in Objective 1.
5. write a general equation to illustrate the preparation of an arylamine by the reduction of a nitro compound, and balance such an equation.
6. identify the product formed from the reduction of a given aromatic nitro compound.
7. identify the organic compound, the inorganic reagents, or both, needed to prepare a given arylamine.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• azide synthesis
• Curtius rearrangement
• Hofmann rearrangement
• Gabriel synthesis
• imide
• reductive amination
Study Notes
You may wish to review the mechanism of SN2 reactions which is discussed in some detail in Sections 11.2 and 11.3.
Azide synthesis is the first method on the table of synthesis of primary amines. The Lewis structure of the azide ion, N3, is as shown below.
An “imide” is a compound in which an N$\ce{-}$H group is attached to two carbonyl groups; that is,
You should note the commonly used trivial names of the following compounds.
The phthalimide alkylation mentioned in the reading is also known as the Gabriel synthesis.
If necessary, review the reduction of nitriles (Section 20.7) and the reduction of amides (Section 21.7).
Before you read the section on reductive amination you may wish to remind yourself of the structure of an imine (see Section 19.8).
The Hofmann rearrangement is usually called the Hofmann degradation. In a true rearrangement reaction, no atoms are lost or gained; however, in this particular reaction one atom of carbon and one atom of oxygen are lost from the amide starting material, thus the term “rearrangement” is not really appropriate. There is a rearrangement step in the overall degradation process, however: this is the step in which the alkyl group of the acyl nitrene migrates from carbon to nitrogen to produce an isocyanate.
You get a complicated series of reactions on heating primary amines with halogenoalkanes to give a mixture of products - probably one of the most confusing sets of reactions you will meet at this level. The products of the reactions include secondary and tertiary amines and their salts, and quaternary ammonium salts.
Making secondary amines and their salts
In the first stage of the reaction, you get the salt of a secondary amine formed. For example if you started with ethylamine and bromoethane, you would get diethylammonium bromide
In the presence of excess ethylamine in the mixture, there is the possibility of a reversible reaction. The ethylamine removes a hydrogen from the diethylammonium ion to give free diethylamine - a secondary amine.
Making tertiary amines and their salts
But it doesn't stop here! The diethylamine also reacts with bromoethane - in the same two stages as before. This is where the reaction would start if you reacted a secondary amine with a halogenoalkane.
In the first stage, you get triethylammonium bromide.
There is again the possibility of a reversible reaction between this salt and excess ethylamine in the mixture.
The ethylamine removes a hydrogen ion from the triethylammonium ion to leave a tertiary amine - triethylamine.
Making a quaternary ammonium salt
The final stage! The triethylamine reacts with bromoethane to give tetraethylammonium bromide - a quaternary ammonium salt (one in which all four hydrogens have been replaced by alkyl groups).
This time there isn't any hydrogen left on the nitrogen to be removed. The reaction stops here.
Preparation of Primary Amines
Although direct alkylation of ammonia by alkyl halides leads to 1º-amines, alternative procedures are preferred in many cases. These methods require two steps, but they provide pure product, usually in good yield. The general strategy is to first form a carbon-nitrogen bond by reacting a nitrogen nucleophile with a carbon electrophile. The following table lists several general examples of this strategy in the rough order of decreasing nucleophilicity of the nitrogen reagent. In the second step, extraneous nitrogen substituents that may have facilitated this bonding are removed to give the amine product.
Nitrogen
Reactant
Carbon
Reactant
1st Reaction
Type
Initial Product
2nd Reaction
Conditions
2nd Reaction
Type
Final Product
N3(–) RCH2-X or
R2CH-X
SN2 RCH2-N3 or
R2CH-N3
LiAlH4 or
4 H2 & Pd
Hydrogenolysis RCH2-NH2 or
R2CH-NH2
C6H5SO2NH(–) RCH2-X or
R2CH-X
SN2 RCH2-NHSO2C6H5 or
R2CH-NHSO2C6H5
Na in NH3 (liq) Hydrogenolysis RCH2-NH2 or
R2CH-NH2
CN(–) RCH2-X or
R2CH-X
SN2 RCH2-CN or
R2CH-CN
LiAlH4 Reduction RCH2-CH2NH2 or
R2CH-CH2NH2
NH3 RCH=O or
R2C=O
Addition /
Elimination
RCH=NH or
R2C=NH
H2 & Ni
or NaBH3CN
Reduction RCH2-NH2 or
R2CH-NH2
NH3 RCOX Addition /
Elimination
RCO-NH2 LiAlH4 Reduction RCH2-NH2
NH2CONH2
(urea)
R3C(+) SN1 R3C-NHCONH2 NaOH soln. Hydrolysis R3C-NH2
A specific example of each general class is provided in the diagram below. In the first two, an anionic nitrogen species undergoes an SN2 reaction with a modestly electrophilic alkyl halide reactant. For example #2 an acidic phthalimide derivative of ammonia has been substituted for the sulfonamide analog listed in the table. The principle is the same for the two cases, as will be noted later. Example #3 is similar in nature, but extends the carbon system by a methylene group (CH2). In all three of these methods 3º-alkyl halides cannot be used because the major reaction path is an E2 elimination.
The methods illustrated by examples #4 and #5 proceed by attack of ammonia, or equivalent nitrogen nucleophiles, at the electrophilic carbon of a carbonyl group. A full discussion of carbonyl chemistry is presented later, but for present purposes it is sufficient to recognize that the C=O double bond is polarized so that the carbon atom is electrophilic. Nucleophile addition to aldehydes and ketones is often catalyzed by acids. Acid halides and anhydrides are even more electrophilic, and do not normally require catalysts to react with nucleophiles. The reaction of ammonia with aldehydes or ketones occurs by a reversible addition-elimination pathway to give imines (compounds having a C=N function). These intermediates are not usually isolated, but are reduced as they are formed (i.e. in situ). Acid chlorides react with ammonia to give amides, also by an addition-elimination path, and these are reduced to amines by LiAlH4.
The 6th example is a specialized procedure for bonding an amino group to a 3º-alkyl group (none of the previous methods accomplishes this). Since a carbocation is the electrophilic species, rather poorly nucleophilic nitrogen reactants can be used. Urea, the diamide of carbonic acid, fits this requirement nicely. The resulting 3º-alkyl-substituted urea is then hydrolyzed to give the amine. One important method of preparing 1º-amines, especially aryl amines, uses a reverse strategy. Here a strongly electrophilic nitrogen species (NO2(+)) bonds to a nucleophilic carbon compound. This nitration reaction gives a nitro group that can be reduced to a 1º-amine by any of several reduction procedures.
The Hofmann rearrangement of 1º-amides provides an additional synthesis of 1º-amines.
Reduction of Nitro Groups
Several methods for reducing nitro groups to amines are known. These include catalytic hydrogenation (H2 + catalyst), zinc or tin in dilute mineral acid, and sodium sulfide in ammonium hydroxide solution. The procedures described above are sufficient for most case
Nitriles can be converted to 1° amines by reaction with LiAlH4
During this reaction the hydride nucleophile attacks the electrophilic carbon in the nitrile to form an imine anion. Once stabilized by a Lewis acid-base complexation the imine salt can accept a second hydride to form a dianion. The dianion can then be converted to an amine by addition of water.
Reductive Amination
Aldehydes and ketones can be converted into 1o, 2o and 3o amines using reductive amination. The reaction takes place in two parts. The first step is the nucleophiic addition of the carbonyl group to form an imine. The second step is the reduction of the imine to an amine using an reducing agent. A reducing agent commonly used for this reaction is sodium cyanoborohydride (NaBH3CN).
Hofmann rearrangement
Hofmann rearrangement, also known as Hofmann degradation and not to be confused with Hofmann elimination, is the reaction of a primary amide with a halogen (chlorine or bromine) in strongly basic (sodium or potassium hydroxide) aqueous medium, which converts the amide to a primary amine. For example:
Mechanism:
Curtius Rearrangement
The Curtius rearrangement involves an acyl azide.
The mechanism of the Curtius rearrangement involves the migration of an -R group form the carbonyl carbon to the the neighboring nitrogen. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/21%3A_Amines_and_Their_Derivatives/21.07%3A_Synthesis_of_Amines__from__Carboxylic_Amides.txt |
Amine functions seldom serve as leaving groups in nucleophilic substitution or base-catalyzed elimination reactions. Indeed, they are even less effective in this role than are hydroxyl and alkoxyl groups. In the case of alcohols and ethers, a useful technique for enhancing the reactivity of the oxygen function was to modify the leaving group (OH(–) or OR(–)) to improve its stability as an anion (or equivalent). This stability is conveniently estimated from the strength of the corresponding conjugate acids.
As noted earlier, 1º and 2º-amines are much weaker acids than alcohols, so it is not surprising that it is difficult to force the nitrogen function to assume the role of a nucleophilic leaving group. For example, heating an amine with HBr or HI does not normally convert it to the corresponding alkyl halide, as in the case of alcohols and ethers. In this context we note that the acidity of the putative ammonium leaving group is at least ten powers of ten less than that of an analogous oxonium species. The loss of nitrogen from diazonium intermediates is a notable exception in this comparison, due to the extreme stability of this leaving group (the conjugate acid of N2 would be an extraordinarily strong acid).
One group of amine derivatives that have proven useful in SN2 and E2 reactions is that composed of the tetraalkyl (4º-) ammonium salts. Most applications involving this class of compounds are eliminations, but a few examples of SN2 substitution have been reported.
C6H5–N(CH3)3(+) Br(–) + R-S(–) Na(+)
acetone & heat
R-S-CH3 + C6H5–N(CH3)2 + NaBr
(CH3)4N(+) OH(–)
heat
CH3–OH + (CH3)3N
Hofmann Elimination
Elimination reactions of 4º-ammonium salts are termed Hofmann eliminations. Since the counter anion in most 4º-ammonium salts is halide, this is often replaced by the more basic hydroxide ion through reaction with silver hydroxide (or silver oxide). The resulting hydroxide salt must then be heated (100 - 200 ºC) to effect the E2-like elimination of a 3º-amine. Example #1 below shows a typical Hofmann elimination. Obviously, for an elimination to occur one of the alkyl substituents on nitrogen must have one or more beta-hydrogens, as noted earlier in examining elimination reactions of alkyl halides.
In example #2 above, two of the alkyl substituents on nitrogen have beta-hydrogens, all of which are on methyl groups (colored orange & magenta). The chief product from the elimination is the alkene having the more highly substituted double bond, reflecting not only the 3:1 numerical advantage of those beta-hydrogens, but also the greater stability of the double bond.
Example #3 illustrates two important features of the Hofmann elimination:
1. Simple amines are easily converted to the necessary 4º-ammonium salts by exhaustive alkylation, usually with methyl iodide (methyl has no beta-hydrogens and cannot compete in the elimination reaction). Exhaustive methylation is shown again in example #4.
2. When a given alkyl group has two different sets of beta-hydrogens available to the elimination process (colored orange & magenta here), the major product is often the alkene isomer having the less substituted double bond.
The tendency of Hofmann eliminations to give the less-substituted double bond isomer is commonly referred to as the Hofmann Rule, and contrasts strikingly with the Zaitsev Rule formulated for dehydrohalogenations and dehydrations. In cases where other activating groups, such as phenyl or carbonyl, are present, the Hofmann Rule may not apply. Thus, if 2-amino-1-phenylpropane is treated in the manner of example #3, the product consists largely of 1-phenylpropene (E & Z-isomers).
To understand why the base-induced elimination of 4º-ammonium salts behaves differently from that of alkyl halides it is necessary to reexamine the nature of the E2 transition state, first described for dehydrohalogenation. The energy diagram shown earlier for a single-step bimolecular E2 mechanism is repeated below.
The E2 transition state is less well defined than is that of SN2 reactions. More bonds are being broken and formed, with the possibility of a continuum of states in which the extent of C–H and C–X bond-breaking and C=C bond-making varies. For example, if the bond to the leaving group (X) is substantially broken relative to the other bond changes, the transition state approaches that for an E1 reaction (initial ionization followed by a fast second step). At the other extreme, if the acidity of the beta-hydrogens is enhanced, then substantial breaking of C–H may occur before the other bonds begin to be affected. For most simple alkyl halides it was proper to envision a balanced transition state, in which there was a synchronous change in all the bonds. Such a model was consistent with the Zaitsev Rule.
When the leaving group X carries a positive charge, as do the 4º-ammonium compounds discussed here, the inductive influence of this charge will increase the acidity of both the alpha and the beta-hydrogens. Furthermore, the 4º-ammonium substituent is much larger than a halide or hydroxyl group and may perturb the conformations available to substituted beta-carbons. It seems that a combination of these factors acts to favor base attack at the least substituted (least hindered and most acidic) set of beta-hydrogens. The favored anti orientation of the leaving group and beta-hydrogen, noted for dehydrohalogenation, is found for many Hofmann eliminations; but syn-elimination is also common, possibly because the attraction of opposite charges orients the hydroxide base near the 4º-ammonium leaving group.
Three additional examples of the Hofmann elimination are shown in the following diagram. Example #1 is interesting in two respects. First, it generates a 4º-ammonium halide salt in a manner different from exhaustive methylation. Second, this salt is not converted to its hydroxide analog prior to elimination. A concentrated aqueous solution of the halide salt is simply dropped into a refluxing sodium hydroxide solution, and the volatile hydrocarbon product is isolated by distillation.
Example #2 illustrates an important aspect of the Hofmann elimination. If the nitrogen atom is part of a ring, then a single application of this elimination procedure does not remove the nitrogen as a separate 3º-amine product. In order to sever the nitrogen function from the molecule, a second Hofmann elimination must be carried out. Indeed, if the nitrogen atom was a member of two rings (fused or spiro), then three repetitions of the Hofmann elimination would be required to sever the nitrogen from the remaining molecular framework.
Example #3 is noteworthy because the less stable trans-cyclooctene is the chief product, accompanied by the cis-isomer. An anti-E2-transition state would necessarily give the cis-cycloalkene, so the trans-isomer must be generated by a syn-elimination. The cis-cyclooctene produced in this reaction could also be formed by a syn-elimination. Cyclooctane is a conformationally complex structure. Several puckered conformations that avoid angle strain are possible, and one of the most stable of these is shown on the right. Some eclipsed bonds occur in all these conformers, and transannular hydrogen crowding is unavoidable. Since the trimethylammonium substituent is large (about the size of tert-butyl) it will probably assume an equatorial-like orientation to avoid steric crowding. An anti-E2 transition state is likely to require an axial-like orientation of this bulky group, making this an unfavorable path. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/21%3A_Amines_and_Their_Derivatives/21.08%3A_Quaternary_Ammonium_Salts%3A__Hofmann_Elimination.txt |
The acid-catalyzed reaction of an enolizable aldehyde or an enolizable ketone with an imminium ion, usually generated in situ by the reaction of formaldehyde with a secondary amine, followed by a base to give a β-aminoaldehyde of a β-aminoketone, respectively, is known as the Mannich reaction. The product of the Mannich reaction is called the Mannich base.
21.10: Nitrosation of Amines
Nitrous acid ($HNO_2$ or $HONO$) reacts with aliphatic amines in a fashion that provides a useful test for distinguishing primary, secondary and tertiary amines.
• 1°-Amines + HONO (cold acidic solution) $\rightarrow$ Nitrogen Gas Evolution from a Clear Solution
• 2°-Amines + HONO (cold acidic solution) $\rightarrow$ An Insoluble Oil (N-Nitrosamine)
• 3°-Amines + HONO (cold acidic solution) $\rightarrow$ A Clear Solution (Ammonium Salt Formation)
Nitrous acid is a Brønsted acid of moderate strength (pKa = 3.3). Because it is unstable, it is prepared immediately before use in the following manner:
Under the acidic conditions of this reaction, all amines undergo reversible salt formation:
This happens with 3º-amines, and the salts are usually soluble in water. The reactions of nitrous acid with 1°- and 2°- aliphatic amines may be explained by considering their behavior with the nitrosonium cation, NO(+), an electrophilic species present in acidic nitrous acid solutions.
Secondary Amines
The distinct behavior of 1º, 2º & 3º-aliphatic amines is an instructive challenge to our understanding of their chemistry, but is of little importance as a synthetic tool. The SN1 product mixtures from 1º-amines are difficult to control, and rearrangement is common when branched primary alkyl groups are involved. The N-nitrosamines formed from 2º-amines are carcinogenic, and are not generally useful as intermediates for subsequent reactions.
Aryl Amines
Nitrous acid reactions of 1º-aryl amines generate relatively stable diazonium species that serve as intermediates for a variety of aromatic substitution reactions. Diazonium cations may be described by resonance contributors, as in the bracketed formulas shown below. The left-hand contributor is dominant because it has greater bonding. Loss of nitrogen is slower than in aliphatic 1º-amines because the C-N bond is stronger, and aryl carbocations are comparatively unstable.
Aqueous solutions of these diazonium ions have sufficient stability at 0º to 10 ºC that they may be used as intermediates in a variety of nucleophilic substitution reactions. For example, if water is the only nucleophile available for reaction, phenols are formed in good yield.
2º-Aryl Amines:
2º-Aryl amines give N-nitrosamine derivatives on reaction with nitrous acid, and thus behave identically to their aliphatic counterparts.
3º-Aryl Amines:
Depending on ring substitution, 3º-Aryl amines may undergo aromatic ring nitrosation at sites ortho or para to the amine substituent. The nitrosonium cation is not sufficiently electrophilic to react with benzene itself, or even toluene, but highly activated aromatic rings such as amines and phenols are capable of substitution. Of course, the rate of reaction of NO(+) directly at nitrogen is greater than that of ring substitution, as shown in the previous example. Once nitrosated, the activating character of the amine nitrogen is greatly diminished; and N-nitrosoaniline derivatives, or indeed any amide derivatives, do not undergo ring nitrosation. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/21%3A_Amines_and_Their_Derivatives/21.09%3A_Mannich___Reaction%3A_Alkylation_of_Enols__by__Iminium__Ions.txt |
Thumbnail: Ball-and-stick model of a phenol molecule. (Public Domain; Benjah-bmm27).
22: Chemistry of the Benzene Substituents: Alkylbenzenes Phenols and Benzenamines
Oxidation of Alkyl Side-Chains
The benzylic hydrogens of alkyl substituents on a benzene ring are activated toward free radical attack, as noted earlier. Furthermore, SN1, SN2 and E1 reactions ofbenzylic halides, show enhanced reactivity, due to the adjacent aromatic ring. The possibility that these observations reflect a general benzylic activation is supported by the susceptibility of alkyl side-chains to oxidative degradation, as shown in the following examples (the oxidized side chain is colored). Such oxidations are normally effected by hot acidic pemanganate solutions, but for large scale industrial operations catalyzed air-oxidations are preferred. Interestingly, if the benzylic position is completely substituted this oxidative degradation does not occur (second equation, the substituted benzylic carbon is colored blue).
C6H5CH2CH2CH2CH3 + KMnO4 + H3O(+) & heat
C6H5CO2H + CO2
p-(CH3)3C–C6H4CH3 + KMnO4 + H3O(+) & heat
p-(CH3)3C–C6H4CO2H
These equations are not balanced. The permanganate oxidant is reduced, usually to Mn(IV) or Mn(II). Two other examples of this reaction are given below, and illustrate its usefulness in preparing substituted benzoic acids.
Reduction of Nitro Groups and Aryl Ketones
Electrophilic nitration and Friedel-Crafts acylation reactions introduce deactivating, meta-directing substituents on an aromatic ring. The attached atoms are in a high oxidation state, and their reduction converts these electron withdrawing functions into electron donating amino and alkyl groups. Reduction is easily achieved either by catalytic hydrogenation (H2 + catalyst), or with reducing metals in acid. Examples of these reductions are shown here, equation 6 demonstrating the simultaneous reduction of both functions. Note that the butylbenzene product in equation 4 cannot be generated by direct Friedel-Crafts alkylation due to carbocation rearrangement. The zinc used in ketone reductions, such as 5, is usually activated by alloying with mercury (a process known as amalgamation).
Several alternative methods for reducing nitro groups to amines are known. These include zinc or tin in dilute mineral acid, and sodium sulfide in ammonium hydroxide solution. The procedures described above are sufficient for most cases.
22.03: Names and Properties of Phenols
Naming phenols
Phenols are named using the rules for aromatic compounds discussed in seciton 15.1 Note! that -phenol is used rather than -benzene.
Compounds like alcohols and phenol which contain an -OH group attached to a hydrocarbon are very weak acids. Alcohols are so weakly acidic that, for normal lab purposes, their acidity can be virtually ignored. However, phenol is sufficiently acidic for it to have recognizably acidic properties - even if it is still a very weak acid. A hydrogen ion can break away from the -OH group and transfer to a base.
Why is phenol acidic?
Compounds like alcohols and phenol which contain an -OH group attached to a hydrocarbon are very weak acids. Alcohols are so weakly acidic that, for normal lab purposes, their acidity can be virtually ignored. However, phenol is sufficiently acidic for it to have recognizably acidic properties - even if it is still a very weak acid. A hydrogen ion can break away from the -OH group and transfer to a base. For example, in solution in water:
Phenol is a very weak acid and the position of equilibrium lies well to the left. Phenol can lose a hydrogen ion because the phenoxide ion formed is stabilised to some extent. The negative charge on the oxygen atom is delocalised around the ring. The more stable the ion is, the more likely it is to form. One of the lone pairs on the oxygen atom overlaps with the delocalised electrons on the benzene ring.
This overlap leads to a delocalization which extends from the ring out over the oxygen atom. As a result, the negative charge is no longer entirely localized on the oxygen, but is spread out around the whole ion.
Spreading the charge around makes the ion more stable than it would be if all the charge remained on the oxygen. However, oxygen is the most electronegative element in the ion and the delocalized electrons will be drawn towards it. That means that there will still be a lot of charge around the oxygen which will tend to attract the hydrogen ion back again. That is why phenol is only a very weak acid.
Why is phenol a much stronger acid than cyclohexanol? To answer this question we must evaluate the manner in which an oxygen substituent interacts with the benzene ring. As noted in our earlier treatment of electrophilic aromatic substitution reactions, an oxygen substituent enhances the reactivity of the ring and favors electrophile attack at ortho and para sites. It was proposed that resonance delocalization of an oxygen non-bonded electron pair into the pi-electron system of the aromatic ring was responsible for this substituent effect. A similar set of resonance structures for the phenolate anion conjugate base appears below the phenol structures.
The resonance stabilization in these two cases is very different. An important principle of resonance is that charge separation diminishes the importance of canonical contributors to the resonance hybrid and reduces the overall stabilization. The contributing structures to the phenol hybrid all suffer charge separation, resulting in very modest stabilization of this compound. On the other hand, the phenolate anion is already charged, and the canonical contributors act to disperse the charge, resulting in a substantial stabilization of this species. The conjugate bases of simple alcohols are not stabilized by charge delocalization, so the acidity of these compounds is similar to that of water. An energy diagram showing the effect of resonance on cyclohexanol and phenol acidities is shown on the right. Since the resonance stabilization of the phenolate conjugate base is much greater than the stabilization of phenol itself, the acidity of phenol relative to cyclohexanol is increased. Supporting evidence that the phenolate negative charge is delocalized on the ortho and para carbons of the benzene ring comes from the influence of electron-withdrawing substituents at those sites.
Properties of phenol as an acid
With indicators
The pH of a typical dilute solution of phenol in water is likely to be around 5 - 6 (depending on its concentration). That means that a very dilute solution isn't really acidic enough to turn litmus paper fully red. Litmus paper is blue at pH 8 and red at pH 5. Anything in between is going to show as some shade of "neutral". Phenol reacts with sodium hydroxide solution to give a colourless solution containing sodium phenoxide.
In this reaction, the hydrogen ion has been removed by the strongly basic hydroxide ion in the sodium hydroxide solution.
With sodium carbonate or sodium hydrogencarbonate
Phenol isn't acidic enough to react with either of these. Or, looked at another way, the carbonate and hydrogencarbonate ions aren't strong enough bases to take a hydrogen ion from the phenol. Unlike the majority of acids, phenol does not give carbon dioxide when you mix it with one of these. This lack of reaction is actually useful. You can recognise phenol because:
• It is fairly insoluble in water.
• It reacts with sodium hydroxide solution to give a colourless solution (and therefore must be acidic).
• It does not react with sodium carbonate or hydrogencarbonate solutions (and so must be only very weakly acidic).
With metallic sodium
Acids react with the more reactive metals to give hydrogen gas. Phenol is no exception - the only difference is the slow reaction because phenol is such a weak acid. Phenol is warmed in a dry tube until it is molten, and a small piece of sodium added. There is some fizzing as hydrogen gas is given off. The mixture left in the tube will contain sodium phenoxide.
Contributors
Jim Clark (Chemguide.co.uk)
Substitution of the hydroxyl hydrogen atom is even more facile with phenols, which are roughly a million times more acidic than equivalent alcohols. This phenolic acidity is further enhanced by electron-withdrawing substituents ortho and para to the hydroxyl group, as displayed in the following diagram. The alcohol cyclohexanol is shown for reference at the top left. It is noteworthy that the influence of a nitro substituent is over ten times stronger in the para-location than it is meta, despite the fact that the latter position is closer to the hydroxyl group. Furthermore additional nitro groups have an additive influence if they are positioned in ortho or para locations. The trinitro compound shown at the lower right is a very strong acid called picric acid.
Why is phenol a much stronger acid than cyclohexanol? To answer this question we must evaluate the manner in which an oxygen substituent interacts with the benzene ring. As noted in our earlier treatment of electrophilic aromatic substitution reactions, an oxygen substituent enhances the reactivity of the ring and favors electrophile attack at ortho and para sites. It was proposed that resonance delocalization of an oxygen non-bonded electron pair into the pi-electron system of the aromatic ring was responsible for this substituent effect. Formulas illustrating this electron delocalization will be displayed when the "Resonance Structures" button beneath the previous diagram is clicked. A similar set of resonance structures for the phenolate anion conjugate base appears below the phenol structures.
The resonance stabilization in these two cases is very different. An important principle of resonance is that charge separation diminishes the importance of canonical contributors to the resonance hybrid and reduces the overall stabilization. The contributing structures to the phenol hybrid all suffer charge separation, resulting in very modest stabilization of this compound. On the other hand, the phenolate anion is already charged, and the canonical contributors act to disperse the charge, resulting in a substantial stabilization of this species. The conjugate bases of simple alcohols are not stabilized by charge delocalization, so the acidity of these compounds is similar to that of water. An energy diagram showing the effect of resonance on cyclohexanol and phenol acidities is shown on the right. Since the resonance stabilization of the phenolate conjugate base is much greater than the stabilization of phenol itself, the acidity of phenol relative to cyclohexanol is increased. Supporting evidence that the phenolate negative charge is delocalized on the ortho and para carbons of the benzene ring comes from the influence of electron-withdrawing substituents at those sites.
Contributors
William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/22%3A_Chemistry_of_the_Benzene_Substituents%3A_Alkylbenzenes_Phenols_and_Benzenamines/22.02%3A_Benzylic__Oxidations_and__Reductions.txt |
Objectives
After completing this section, you should be able to
1. identify the conditions necessary for an aryl halide to undergo nucleophilic aromatic substitution, and give an example of such a reaction.
2. write the detailed mechanism for a nucleophilic aromatic substitution reaction.
3. compare the mechanism of a nucleophilic aromatic substitution reaction and the SN1 and SN2 mechanisms discussed earlier.
4. identify the product formed when a given nucleophile reacts with a given aryl halide in a nucleophilic aromatic substitution reaction.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• Meisenheimer complex
• nucleophilic aromatic substitution
Study Notes
A nucleophilic aromatic substitution reaction is a reaction in which one of the substituents in an aromatic ring is replaced by a nucleophile.
A Meisenheimer complex is a negatively charged intermediate formed by the attack of a nucleophile upon one of the aromatic-ring carbons during the course of a nucleophilic aromatic substitution reaction. A typical Meisenheimer complex is shown in the reaction scheme below. Notice how this particular complex can be formed from two different starting materials by using a different nucleophile in each case.
Figure 16.2: The formation of a typical Meisenheimer complex
A Nucleophilic Aromatic Displacement Reactions of Aryl Halides
The carbon-halogen bonds of aryl halides are like those of alkenyl halides in being much stronger than those of alkyl halides (see Table 4-6). The simple aryl halides generally are resistant to attack by nucleophiles in either SN1 or SN2 reactions (Table 14-6). However, this low reactivity can be changed dramatically by changes in the reaction conditions and the structure of the aryl halide. In fact, nucleophilic displacement becomes quite rapid
1. when the aryl halide is activated by substitution with strongly electron-attracting groups such as NO2, and
2. when very strongly basic nucleophilic reagents are used.
Addition-Elimination Mechanism of Nucleophilic Substitution of Aryl Halides
Although the simple aryl halides are inert to the usual nucleophilic reagents, considerable activation is produced by strongly electron-attracting substituents provided these are located in either the ortho or para positions, or both. For example, the displacement of chloride ion from 1-chloro-2,4-dinitrobenzene by dimethylamine occurs readily in ethanol solution at room temperature. Under the same conditions chlorobenzene completely fails to react; thus the activating influence of the two nitro groups amounts to a factor of at least 108:
A related reaction is that of 2,4-dinitrofluorobenzene with the amino groups of peptides and proteins, and this reaction provides a means for analysis of the N-terminal amino acids in polypeptide chains. (See Section 25-7B.)
In general, the reactions of activated aryl halides closely resemble the SN2-displacement reactions of aliphatic halides. The same nucleophilic reagents are effective (e.g., CH3O⊖, HO⊖, and RNH2); the reactions are second order overall (first order in halide and first order in nucleophile); and for a given halide the more nucleophilic the attacking reagent, the faster the reaction. However, there must be more than a subtle difference in mechanism because an aryl halide is unable, to pass through the same type of transition state as an alkyl halide in SN2 displacements.
The generally accepted mechanism of nucleophilic aromatic substitution of aryl halides carrying activating groups involves two steps that are closely analogous to those briefly described in Section 14-4 for alkenyl and alkynyl halides. The first step involves attack of the nucleophile Y:⊖ at the carbon bearing the halogen substituent to form an intermediate carbanion 4 (Equation 14-3). The aromatic system is destroyed on forming the anion, and the carbon at the reaction site changes from planar (sp2 bonds) to tetrahedral (sp3 bonds).
In the second step, loss of an anion, X⊖ or Y⊖, regenerates an aromatic system, and, if X⊖ is lost, the overall reaction is nucleophilic displacement of X by Y (Equation 14-4).
In the case of a neutral nucleophilic reagent, Y or HY, the reaction sequence would be the same except for the necessary adjustments in the charge of the intermediate:
Why is this reaction pathway generally unfavorable for the simple aryl halides? The answer is that the intermediate 4, which we can express as a hybrid of the valence-bond structures 4a-4c, is too high in energy to be formed at any practical rate. Not only has 4 lost the aromatic stabilization of the benzene ring, but its formation results in transfer of negative charge to the ring carbons, which themselves are not very electronegative:
However, when strongly electron-attracting groups are located on the ring at the ortho-para positions, the intermediate anion is stabilized by delocalization of electrons from the ring carbons to more favorable locations on the substituent groups. As an example, consider the displacement of bromine by OCH3 in the reaction of 4-bromonitrobenzene and methoxide ion:
The anionic intermediate formed by addition of methoxide ion to the aryl halide can be described by the valence-bond structures 5a-5d. Of these structures 5d is especially important because in it the charge is transferred from the ring carbons to the oxygen of the nitro substituent:
Substituents in the meta positions have much less effect on the reactivity of an aryl halide because delocalization of electrons to the substituent is not possible. No formulas can be written analogous to 5c and 5d in which the negative charges are both on atoms next to positive nitrogen, and
In a few instances, stable compounds resembling the postulated reaction intermediate have been isolated. One classic example is the complex 7 (isolated by J. Meisenheimer), which is the product of the reaction of either the methyl aryl ether 6 with potassium ethoxide, or the ethyl aryl ether 8 and potassium methoxide:
Contributors and Attributions
John D. Robert and Marjorie C. Caserio (1977) Basic Principles of Organic Chemistry, second edition. W. A. Benjamin, Inc. , Menlo Park, CA. ISBN 0-8053-8329-8. This content is copyrighted under the following conditions, "You are granted permission for individual, educational, research and non-commercial reproduction, distribution, display and performance of this work in any format."
Objectives
After completing this section, you should be able to
1. identify the reagents and conditions required to produce phenol from chlorobenzene on an industrial scale.
2. write the mechanism for the conversion of an alkyl halide to a phenol through a benzyne intermediate.
3. discuss the experimental evidence which supports the existence of benzyne intermediates.
4. discuss the bonding in benzyne, and hence account for its high reactivity.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• benzyne
• elimination-addition mechanism
Study Notes
An elimination-addition mechanism involves the elimination of the elements of a small molecule from a substrate to produce a highly reactive intermediate, which then undergoes an addition reaction.
The elimination-addition mechanism of nucleophilic aromatic substitution involves the remarkable intermediate called benzyne or arynes.
14-6C Elimination-Addition Mechanism of Nucleophilic Aromatic Substitution. Arynes
The reactivities of aryl halides, such as the halobenzenes, are exceedingly low toward nucleophilic reagents that normally effect displacements with alkyl halides and activated aryl halides. Substitutions do occur under forcing conditions of either high temperatures or very strong bases. For example, chlorobenzene reacts with sodium hydroxide solution at temperatures around $340^\text{o}$ and this reaction was once an important commercial process for the production of benzenol (phenol):
In addition, aryl chlorides, bromides, and iodides can be converted to areneamines $\ce{ArNH_2}$ by the conjugate bases of amines. In fact, the reaction of potassium amide with bromobenzene is extremely rapid, even at temperatures as low as $-33^\text{o}$ with liquid ammonia as solvent:
However, displacement reactions of this type differ from the previously discussed displacements of activated aryl halides in that rearrangement often occurs. That is, the entering group does not always occupy the same position on the ring as that vacated by the halogen substituent. For example, the hydrolysis of 4-chloromethylbenzene at $340^\text{o}$ gives an equimolar mixture of 3- and 4-methylbenzenols:
Even more striking is the exclusive formation of 3-methoxybenzenamine in the amination of 2-chloromethoxybenzene. Notice that this result is a violation of the principle of least structural change (Section 1-1H):
The mechanism of this type of reaction has been studied extensively, and much evidence has accumulated in support of a stepwise process, which proceeds first by base-catalyzed elimination of hydrogen halide $\left( \ce{HX} \right)$ from the aryl halide - as illustrated below for the amination of bromobenzene:
Elimination
The product of the elimination reaction is a highly reactive intermediate $9$ called benzyne, or dehydrobenzene, which differs from benzene in having two less hydrogen and an extra bond between two ortho carbons. Benzyne reacts rapidly with any available nucleophile, in this case the solvent, ammonia, to give an addition product:
Addition
The rearrangements in these reactions result from the attack of the nucleophile at one or the other of the carbons of the extra bond in the intermediate. With benzyne the symmetry is such that no rearrangement would be detected. With substituted benzynes isomeric products may result. Thus 4-methylbenzyne, $10$, from the reaction of hydroxide ion with 4-chloro-1-methylbenzene gives both 3- and 4-methylbenzenols:
In the foregoing benzyne reactions the base that produces the benzyne in the elimination step is derived from the nucleophile that adds in the addition step. This need not always be so, depending on the reaction conditions. In fact, the synthetic utility of aryne reactions depends in large part of the success with which the aryne can be generated by one reagent but captured by another. One such method will be discussed in Section 14-10C and involves organometallic compounds derived from aryl halides. Another method is to generate the aryne by thermal decomposition of a 1,2-disubstituted arene compound such as $11$, in which both substituents are leaving groups - one leaving with an electron pair, the other leaving without:
When $11$ decomposes in the presence of an added nucleophile, the benzyne intermediate is trapped by the nucleophile as it is formed. Or, if a conjugated diene is present, benzyne will react with it by a [4 + 2] cycloaddition. In the absence of other compounds with which it can react, benzyne will undergo [2 + 2] cycloaddition to itself:
Contributors and Attributions
John D. Robert and Marjorie C. Caserio (1977) Basic Principles of Organic Chemistry, second edition. W. A. Benjamin, Inc. , Menlo Park, CA. ISBN 0-8053-8329-8. This content is copyrighted under the following conditions, "You are granted permission for individual, educational, research and non-commercial reproduction, distribution, display and performance of this work in any format." | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/22%3A_Chemistry_of_the_Benzene_Substituents%3A_Alkylbenzenes_Phenols_and_Benzenamines/22.04%3A_Preparation__of_Phenols%3A__Nucleophilic__Aromatic__Substitut.txt |
As with the alcohols, the phenolic hydroxyl hydrogen is rather easily replaced by other substituents. For example, phenol reacts easily with acetic anhydride to give phenyl acetate. Likewise, the phenolate anion is an effective nucleophile in SN2 reactions, as in the second example below.
Examples:
Contributors
Prof. Steven Farmer (Sonoma State University)
William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry
22.06: Electrophilic Substitution of Phenols
Over reaction of Aniline and Phenol
The strongest activating and ortho/para-directing substituents are the amino (-NH2) and hydroxyl (-OH) groups. Direct nitration of phenol (hydroxybenzene) by dilute nitric acid gives modest yields of nitrated phenols and considerable oxidative decomposition to tarry materials; aniline (aminobenzene) is largely destroyed. Bromination of both phenol and aniline is difficult to control, with di- and tri-bromo products forming readily. Because of their high nucleophilic reactivity, aniline and phenol undergo substitution reactions with iodine, a halogen that is normally unreactive with benzene derivatives. The mixed halogen iodine chloride (ICl) provides a more electrophilic iodine moiety, and is effective in iodinating aromatic rings having less powerful activating substituents.
C6H5–NH2 + I2 + NaHCO3 p-I–C6H4–NH2 + NaI + CO2 + H2O
By acetylating the heteroatom substituent on phenol and aniline, its activating influence can be substantially attenuated. For example, acetylation of aniline gives acetanilide (first step in the following equation), which undergoes nitration at low temperature, yielding the para-nitro product in high yield. The modifying acetyl group can then be removed by acid-catalyzed hydrolysis (last step), to yield para-nitroaniline. Although the activating influence of the amino group has been reduced by this procedure, the acetyl derivative remains an ortho/para-directing and activating substituent.
C6H5–NH2 + (CH3CO)2O
pyridine (a base)
C6H5–NHCOCH3
HNO3 , 5 ºC
p-O2N–C6H4–NHCOCH3
H3O(+) & heat
p-O2N–C6H4–NH
The following diagram illustrates how the acetyl group acts to attenuate the overall electron donating character of oxygen and nitrogen. The non-bonding valence electron pairs that are responsible for the high reactivity of these compounds (blue arrows) are diverted to the adjacent carbonyl group (green arrows). However, the overall influence of the modified substituent is still activating and ortho/para-directing.
Objectives
After completing this section, you should be able to:
1. explain why phenols and phenoxide ions are very reactive towards electrophilic aromatic substitution (see Section 16.4 of the textbook).
2. write an equation to illustrate the oxidation of a phenol or an arylamine to a quinone, and identify the reagents used to oxidize phenols.
3. write an equation to illustrate the reduction of a quinone to a hydroquinone, and identify the reagents used to reduce quinones.
4. describe, briefly, the biological importance of the redox properties of quinones.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• hydroquinone
• quinone
• ubiquinone
Study Notes
“Quinone” is a term used to describe cyclohexadiendiones in general, and p‑benzoquinone in particular. In addition to benzene, other aromatic systems also give rise to quinones; for example, 1,4‑naphthoquinone
“Hydroquinones” are produced by the reduction of quinones according to the following half‑reaction:
“Ubiquinones” are naturally occurring quinones whose role is to transfer a pair of electrons from one substance to another in enzyme‑catalyzed reactions. Ubiquinones are also called coenzymes Q.
Electrophilic Aromatic Substitution Reactions
The facility with which the aromatic ring of phenols and phenol ethers undergoes electrophilic substitution has been noted. Two examples are shown in the following diagram. The first shows the Friedel-Crafts synthesis of the food preservative BHT from para-cresol. The second reaction is interesting in that it further demonstrates the delocalization of charge that occurs in the phenolate anion. Carbon dioxide is a weak electrophile and normally does not react with aromatic compounds; however, the negative charge concentration on the phenolate ring enables the carboxylation reaction shown in the second step. The sodium salt of salicylic acid is the major product, and the preference for ortho substitution may reflect the influence of the sodium cation. This is called the Kolbe-Schmidt reaction, and it has served in the preparation of aspirin, as the last step illustrates.
Oxidation of Phenols: Quinones
Phenols are rather easily oxidized despite the absence of a hydrogen atom on the hydroxyl bearing carbon. Among the colored products from the oxidation of phenol by chromic acid is the dicarbonyl compound para-benzoquinone (also known as 1,4-benzoquinone or simply quinone); an ortho isomer is also known. These compounds are easily reduced to their dihydroxybenzene analogs, and it is from these compounds that quinones are best prepared. Note that meta-quinones having similar structures do not exist. The redox equilibria between the dihydroxybenzenes hydroquinone and catechol and their quinone oxidation states are so facile that milder oxidants than chromate (Jones reagent) are generally preferred.
One such oxidant is Fremy's salt, shown on the right. Reducing agents other than stannous chloride (e.g. NaBH4) may be used for the reverse reaction. The position of the quinone-hydroquinone redox equilibrium is proportional to the square of the hydrogen ion concentration, as shown by the following half-reactions (electrons are colored blue). The electrode potential for this interconversion may therefore be used to measure the pH of solutions.
Quinone + 2H(+)
2e(–)
–2e(–)
Hydroquinone | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/22%3A_Chemistry_of_the_Benzene_Substituents%3A_Alkylbenzenes_Phenols_and_Benzenamines/22.05%3A_Alcohol__Chemistry_of_Phenols.txt |
Objectives
After completing this section, you should be able to
1. write an equation to represent the Claisen rearrangement of allyl phenyl ester.
2. account for the formation of a specific product from a Claisen rearrangement, without giving mechanistic details.
Key Terms
Make certain that you can define, and use in context, the key term below.
Claisen Rearrangements
A Claisen rearrangement is a reaction which is specific to ally aryl ethers and allyl vinyl ethers. Heating an allyl aryl ether to 250 oC causes an intramolecular rearrangement to produce an o-allylphenol.
The Claisen rearrangement takes place through a concerted mechanism in which a C-C bond forms between the C3 of the allyl group and the ortho position of the benzene ring at the same time that the C-O bond of the ether breaks. This rearrangement initially produces the non-aromatic 6-allyl-2,4- cyclohexadienone intermediate which quickly undergoes a proton shift to reform the aromatic ring in the o-allylphenol product. Claisen rearrangement occurs in a six-membered, cyclic transition state involving the concerted movement of six bonding electrons in the first step. The presence of six electrons in a ring suggests that the transition state may have aromatic characteristics. The Claisen rearrangement is part of a broader class of reactions called sigmatropic rearrangements which will be discussed in more detail in Section 30-8.
Evidence for this mechanism was provided by performing the rearrangement with allyl group with a 14C label at C3. The product of this reaction was shown to have the 14C labeled carbon exclusively bonded to the ring.
Allyl vinyl ethers can also undergo a Claisen rearrangement when heated to form gamma, delta -unsaturated ketones or aldehydes.
Claisen rearrangements are rare in biological chemistry. One example is the chorismate mutase catalyzed Claisen rearrangement of chorismate (a allylic vinyl ether) to form prephenate. Prephenate is a precursor in the biosynthetic pathway of aromatic amino acids phenylalanine and tyrosine.
.
Exercise \(1\)
1) Show how you could synthesize allyl phenyl ether from allyl bromide and phenol.
2) What would be the expected product of the following Claisen rearrangement?
Answer
1)
2)
22.08: Oxidation of Phenols: Benzoquinones
Phenols are rather easily oxidized despite the absence of a hydrogen atom on the hydroxyl bearing carbon. Among the colored products from the oxidation of phenol by chromic acid is the dicarbonyl compound para-benzoquinone (also known as 1,4-benzoquinone or simply quinone); an ortho isomer is also known. These compounds are easily reduced to their dihydroxybenzene analogs, and it is from these compounds that quinones are best prepared. Note that meta-quinones having similar structures do not exist. The redox equilibria between the dihydroxybenzenes hydroquinone and catechol and their quinone oxidation states are so facile that milder oxidants than chromate (Jones reagent) are generally preferred. One such oxidant is Fremy's salt, shown on the right. Reducing agents other than stannous chloride (e.g. NaBH4) may be used for the reverse reaction.
The position of the quinone-hydroquinone redox equilibrium is proportional to the square of the hydrogen ion concentration, as shown by the following half-reactions (electrons are colored blue). The electrode potential for this interconversion may therefore be used to measure the pH of solutions.
Quinone + 2H(+)
2e(–)
–2e(–)
Hydroquinone
Further Reading
Chemgapedia
Carey 5th Ed Online | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/22%3A_Chemistry_of_the_Benzene_Substituents%3A_Alkylbenzenes_Phenols_and_Benzenamines/22.07%3A_An__Electrocyclic_Reaction__of__the_Benzene__Ring%3A_The__Cla.txt |
Objectives
After completing this section, you should be able to
1. identify the product formed when a given arylamine is reacted with aqueous bromine.
2. give an appropriate example to illustrate the high reactivity of arylamines in electrophilic aromatic substitution reactions.
3. explain why arylamines cannot be used in Friedel‑Crafts reactions.
1. show how the problems associated with carrying out electrophilic aromatic substitution reactions on arylamines can be circumvented by first converting the amine to an amide, and illustrate this process with an appropriate example.
2. outline a possible synthetic route for the preparation of a given sulfa drug.
3. identify the starting material, the necessary organic reagents, inorganic reagents, or both, and the intermediate compounds formed during the synthesis of a given sulfa drug.
4. design a multi‑step synthesis for a given compound in which it is necessary to protect the amino group by acetylation.
1. write a general equation to describe the formation of an arenediazonium salt.
2. identify the product formed when a given arenediazonium salt is reacted with any of the following compounds: copper(I) chloride, copper(I) bromide, sodium iodide, copper(I) cyanide, hot aqueous acid, hypophosphorous acid.
3. identify the arenediazonium salt, the inorganic reagents, or both, needed to produce a given compound by a diazonium replacement reaction.
1. illustrate, with appropriate examples, the importance to the synthetic chemist of the overall reaction sequence A nitration, B reduction, C diazotization, and D replacement.
2. show how the removal of an amino (or nitro) group from an aromatic ring through the reaction of an arenediazonium salt with hypophosphorous acid (H3PO2) can sometimes be of use in organic synthesis.
1. write a general equation to represent a diazonium coupling reaction.
2. write the detailed mechanism for the coupling reaction which takes place between arenediazonium salts and the electon‑rich aromatic rings of phenols and arylamines.
3. identify the product formed from the reaction of a given arenediazonium salt with a given arylamine or phenol.
4. identify the arenediazonium salt and the arylamine or phenol needed to prepare a given azo compound.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• arenediazonium salt
• azo compound
• Sandmeyer reaction
• sulfa drug
Study Notes
This section contains a considerable amount of new information. To absorb all of it, you should use the three subsections indicated in the reading: electrophilic aromatic substitution and overreaction of aniline (Objectives 1 and 2), the preparation of diazonium salts and the Sandmeyer reaction (Objectives 3 and 4), and diazonium coupling reactions (Objective 5).
The general process in which an aromatic amine is reacted with acetic anhydride, substituted, and then hydrolyzed is known as “protecting the amino group.”
Sulfa drugs have the general formula
Typical examples are sulfathiazole
and sulfapyridine
The reagent used to bring about the chlorosulfonation of acetanilide is chlorosulfuric acid
Sulfanilamide itself is toxic to humans, but derivatives of this compound, such as sulfathiazole, are less harmful to humans and are effective in killing many types of bacteria. The drugs work by “deceiving” the bacteria in the following way. To survive, many micro‑organisms use p‑aminobenzoic acid to synthesize folic acid, a coenzyme in a number of biochemical processes. These micro‑organisms cannot distinguish between sulfa drugs and p‑aminobenzoic acid; so, when the drug is administered, the bacteria use it to produce a compound which has a structure similar to that of folic acid, but which is unable to act as a coenzyme in essential biochemical processes. The result is that many of the bacteria’s amino acids and nucleotides cannot be made, and the bacteria die. Amino acids are discussed in Chapter 26; nucleotides are discussed in Section 28.1.
An “arenediazonium salt” is formed by the reaction of an aromatic amine with nitrous acid at 0–5°C, and has the structure shown below.
Alkanediazonium salts are very unstable; therefore, arenediazonium salts are often simply referred to as diazonium salts.
As is mentioned in the textbook, arenediazonium salts are very useful intermediates from which a wide variety of aromatic compounds can be prepared. You should be thoroughly familiar with the use of diazonium salts to prepare each of the classes of compounds. In addition, you should be aware that fluoroarenes can also be prepared from diazonium salts, as follows:
In this case the diazonium salt is prepared using fluoroboric acid, HBF4, and sodium nitrite. The thermal decomposition of the salt, called the Schiemann reaction, can be quite hazardous.
Note that the IUPAC‑preferred name for cuprous chloride is copper(I) chloride; similarly, cuprous cyanide is called copper(I) cyanide.
Hypophosphorous acid has the structure shown below:
The presence of the letters “azo” in a compound’s name usually implies that a nitrogen‑nitrogen double bond is present in its structure. Azo compounds t in which two aryl groups are joined by an $\ce{-}$N$\ce{=}$N$\ce{-}$ linkage are usually very colourful.
Overreaction of Aniline
Arylamines are very reactive towards electrophilic aromomatic substitution. The strongest activating and ortho/para-directing substituents are the amino (-NH2) and hydroxyl (-OH) groups. Direct nitration of phenol (hydroxybenzene) by dilute nitric acid gives modest yields of nitrated phenols and considerable oxidative decomposition to tarry materials; aniline (aminobenzene) is largely destroyed. Monobromination of both phenol and aniline is difficult to control, with di- and tri-bromo products forming readily.
Because of their high nucleophilic reactivity, aniline and phenol undergo substitution reactions with iodine, a halogen that is normally unreactive with benzene derivatives. The mixed halogen iodine chloride (ICl) provides a more electrophilic iodine moiety, and is effective in iodinating aromatic rings having less powerful activating substituents.
C6H5–NH2 + I2 + NaHCO3 p-I–C6H4–NH2 + NaI + CO2 + H2O
In addition to overreactivity, we have previously seen (Section 16.3) that Friedel-Crafts reactions employing AlCl3 catalyst does not work with aniline. A salt complex forms and prevents electrophilic aromatic substitution.
Both this problem and the aniline overreactivity can be circumvented by first going through the corresponding amide.
Modifying the Influence of Strong Activating Groups
By acetylating the heteroatom substituent on aniline, its activating influence can be substantially attenuated. For example, acetylation of aniline gives acetanilide (first step in the following equation), which undergoes nitration at low temperature, yielding the para-nitro product in high yield. The modifying acetyl group can then be removed by acid-catalyzed hydrolysis (last step), to yield para-nitroaniline. Although the activating influence of the amino group has been reduced by this procedure, the acetyl derivative remains an ortho/para-directing and activating substituent.
C6H5–NH2 + (CH3CO)2O
pyridine (a base)
C6H5–NHCOCH3
HNO3 , 5 ºC
p-O2N–C6H4–NHCOCH3
H3O(+) & heat
p-O2N–C6H4–NH
The following diagram illustrates how the acetyl group acts to attenuate the overall electron donating character of oxygen and nitrogen. The non-bonding valence electron pairs that are responsible for the high reactivity of these compounds (blue arrows) are diverted to the adjacent carbonyl group (green arrows). However, the overall influence of the modified substituent is still activating and ortho/para-directing.
.
Sulfa Drug Synthesis
Sulfa drugs are an important group of synthetic antimicrobial agents (pharmaceuticals) that contain the sulfonamide group. The synthesis of sulfanilamide (a sulfa drug) illustrates how the reactivity of aniline can be modified to make possible an electrophilic aromatic substitution. The corresponding acetanilide undergoes chlorosulfonation. The resulting 4-acetamidobenzenesulfanyl chloride is treated with ammonia to replace the chlorine with an amino group and affords 4-acetamidobenzenesulfonamide. The subsequent hydrolysis of the sulfonamide produces the sulfanilamide.
Diazonium Salts: The Sandmeyer Reaction
Aryl diazonium salts are important intermediates. They are prepared in cold (0 º to 10 ºC) aqueous solution, and generally react with nucleophiles with loss of nitrogen. Some of the more commonly used substitution reactions are shown in the following diagram. Since the leaving group (N2) is thermodynamically very stable, these reactions are energetically favored. Those substitution reactions that are catalyzed by cuprous salts are known as Sandmeyer reactions. Fluoride substitution occurs on treatment with BF4(–), a reaction known as the Schiemann reaction. Stable diazonium tetrafluoroborate salts may be isolated, and on heating these lose nitrogen to give an arylfluoride product. The top reaction with hypophosphorus acid, H3PO2, is noteworthy because it achieves the reductive removal of an amino (or nitro) group. Unlike the nucleophilic substitution reactions, this reduction probably proceeds by a radical mechanism.
These aryl diazonium substitution reactions significantly expand the tactics available for the synthesis of polysubstituted benzene derivatives. Consider the following options:
1. The usual precursor to an aryl amine is the corresponding nitro compound. A nitro substituent deactivates an aromatic ring and directs electrophilic substitution to meta locations.
2. Reduction of a nitro group to an amine may be achieved in several ways. The resulting amine substituent strongly activates an aromatic ring and directs electrophilic substitution to ortho & para locations.
3. The activating character of an amine substituent may be attenuated by formation of an amide derivative (reversible), or even changed to deactivating and meta-directing by formation of a quaternary-ammonium salt (irreversible).
4. Conversion of an aryl amine to a diazonium ion intermediate allows it to be replaced by a variety of different groups (including hydrogen), which may in turn be used in subsequent reactions.
The following examples illustrate some combined applications of these options to specific cases. You should try to conceive a plausible reaction sequence for each. Once you have done so, you may check suggested answers below.
Answer 1:
It should be clear that the methyl substituent will eventually be oxidized to a carboxylic acid function. The timing is important, since a methyl substituent is ortho/para-directing and the carboxyl substituent is meta-directing. The cyano group will be introduced by a diazonium intermediate, so a nitration followed by reduction to an amine must precede this step.
Answer 2:
The hydroxyl group is a strong activating substituent and would direct aromatic ring chlorination to locations ortho & para to itself, leading to the wrong product. As an alternative, the nitro group is not only meta-directing, it can be converted to a hydroxyl group by way of a diazonium intermediate. The resulting strategy is self evident.
Answer 3:
Selective introduction of a fluorine is best achieved by treating a diazonium intermediate with boron tetrafluoride anion. To get the necessary intermediate we need to make p-nitroaniline. Since the nitro substituent on the starting material would direct a new substituent to a meta-location, we must first reduce it to an ortho/para-directing amino group. Amino groups are powerful activating substituents, so we deactivate it by acetylation before nitration. The acetyl substituent also protects the initial amine function from reaction with nitrous acid later on. It is removed in the last step.
Answer 4:
Polybromination of benzene would lead to ortho/para substitution. In order to achieve the mutual meta-relationship of three bromines, it is necessary to introduce a powerful ortho/para-directing prior to bromination, and then remove it following the tribromination. An amino group is ideal for this purpose. Reductive removal of the diazonium group may be accomplished in several ways (three are shown).
Answer 5:
The propyl substituent is best introduced by Friedel-Crafts acylation followed by reduction, and this cannot be carried out in the presence of a nitro substituent. Since an acyl substituent is a meta-director, it is logical to use this property to locate the nitro and chloro groups before reducing the carbonyl moiety. The same reduction method can be used to reduce both the nitro group (to an amine) and the carbonyl group to propyl. We have already seen the use of diazonium intermediates as precursors to phenols.
Answer 6:
Aromatic iodination can only be accomplished directly on highly activated benzene compounds, such as aniline, or indirectly by way of a diazonium intermediate. Once again, a deactivated amino group is the precursor of p-nitroaniline (prb.#3). This aniline derivative requires the more electrophilic iodine chloride (ICl) for ortho-iodination because of the presence of a deactivating nitro substituent. Finally, the third iodine is introduced by the diazonium ion procedure.
Diazonium Coupling Reactions
A resonance description of diazonium ions shows that the positive charge is delocalized over the two nitrogen atoms. It is not possible for nucleophiles to bond to the inner nitrogen, but bonding (or coupling) of negative nucleophiles to the terminal nitrogen gives neutral azo compounds. As shown in the following equation, this coupling to the terminal nitrogen should be relatively fast and reversible. The azo products may exist as E / Z stereoisomers. In practice it is found that the E-isomer predominates at equilibrium.
Unless these azo products are trapped or stabilized in some manner, reversal to the diazonium ion and slow nucleophilic substitution at carbon (with irreversible nitrogen loss) will be the ultimate course of reaction, as described in the previous section. For example, if phenyldiazonium bisufate is added rapidly to a cold solution of sodium hydroxide a relatively stable solution of sodium phenyldiazoate (the conjugate base of the initially formed diazoic acid) is obtained. Lowering the pH of this solution regenerates phenyldiazoic acid (pKa ca. 7), which disassociates back to the diazonium ion and eventually undergoes substitution, generating phenol.
C6H5N2(+) HSO4(–) + NaOH (cold solution) C6H5N2–OH + NaOH (cold) C6H5N2–O(–) Na(+)
phenyldiazonium bisulfate phenyldiazoic acid sodium phenyldiazoate
Aryl diazonium salts may be reduced to the corresponding hydrazines by mild reducing agents such as sodium bisulfite, stannous chloride or zinc dust. The bisulfite reduction may proceed by an initial sulfur-nitrogen coupling, as shown in the following equation.
Ar-N2(+) X(–)
NaHSO3
Ar-N=N-SO3H
NaHSO3
Ar-NH-NH-SO3H
H2O
Ar-NH-NH2 + H2SO4
The most important application of diazo coupling reactions is electrophilic aromatic substitution of activated benzene derivatives by diazonium electrophiles. The products of such reactions are highly colored aromatic azo compounds that find use as synthetic dyestuffs, commonly referred to as azo dyes. Azobenzene (Y=Z=H) is light orange; however, the color of other azo compounds may range from red to deep blue depending on the nature of the aromatic rings and the substituents they carry. Azo compounds may exist as cis/trans isomer pairs, but most of the well-characterized and stable compounds are trans.
Some examples of azo coupling reactions are shown below. A few simple rules are helpful in predicting the course of such reactions:
1. At acid pH (< 6) an amino group is a stronger activating substituent than a hydroxyl group (i.e. a phenol). At alkaline pH (> 7.5) phenolic functions are stronger activators, due to increased phenoxide base concentration.
2. Coupling to an activated benzene ring occurs preferentially para to the activating group if that location is free. Otherwise ortho-coupling will occur.
3. Naphthalene normally undergoes electrophilic substitution at an alpha-location more rapidly than at beta-sites; however, ortho-coupling is preferred. See the diagram for examples of α / β notation in naphthalenes.
You should try to conceive a plausible product structure for each of the following couplings. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/22%3A_Chemistry_of_the_Benzene_Substituents%3A_Alkylbenzenes_Phenols_and_Benzenamines/22.10%3A_Arenediazonium_Salts.txt |
Objectives
After completing this section, you should be able to
1. identify the product formed when a given arylamine is reacted with aqueous bromine.
2. give an appropriate example to illustrate the high reactivity of arylamines in electrophilic aromatic substitution reactions.
3. explain why arylamines cannot be used in Friedel‑Crafts reactions.
1. show how the problems associated with carrying out electrophilic aromatic substitution reactions on arylamines can be circumvented by first converting the amine to an amide, and illustrate this process with an appropriate example.
2. outline a possible synthetic route for the preparation of a given sulfa drug.
3. identify the starting material, the necessary organic reagents, inorganic reagents, or both, and the intermediate compounds formed during the synthesis of a given sulfa drug.
4. design a multi‑step synthesis for a given compound in which it is necessary to protect the amino group by acetylation.
1. write a general equation to describe the formation of an arenediazonium salt.
2. identify the product formed when a given arenediazonium salt is reacted with any of the following compounds: copper(I) chloride, copper(I) bromide, sodium iodide, copper(I) cyanide, hot aqueous acid, hypophosphorous acid.
3. identify the arenediazonium salt, the inorganic reagents, or both, needed to produce a given compound by a diazonium replacement reaction.
1. illustrate, with appropriate examples, the importance to the synthetic chemist of the overall reaction sequence A nitration, B reduction, C diazotization, and D replacement.
2. show how the removal of an amino (or nitro) group from an aromatic ring through the reaction of an arenediazonium salt with hypophosphorous acid (H3PO2) can sometimes be of use in organic synthesis.
1. write a general equation to represent a diazonium coupling reaction.
2. write the detailed mechanism for the coupling reaction which takes place between arenediazonium salts and the electon‑rich aromatic rings of phenols and arylamines.
3. identify the product formed from the reaction of a given arenediazonium salt with a given arylamine or phenol.
4. identify the arenediazonium salt and the arylamine or phenol needed to prepare a given azo compound.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• arenediazonium salt
• azo compound
• Sandmeyer reaction
• sulfa drug
Study Notes
This section contains a considerable amount of new information. To absorb all of it, you should use the three subsections indicated in the reading: electrophilic aromatic substitution and overreaction of aniline (Objectives 1 and 2), the preparation of diazonium salts and the Sandmeyer reaction (Objectives 3 and 4), and diazonium coupling reactions (Objective 5).
The general process in which an aromatic amine is reacted with acetic anhydride, substituted, and then hydrolyzed is known as “protecting the amino group.”
Sulfa drugs have the general formula
Typical examples are sulfathiazole
and sulfapyridine
The reagent used to bring about the chlorosulfonation of acetanilide is chlorosulfuric acid
Sulfanilamide itself is toxic to humans, but derivatives of this compound, such as sulfathiazole, are less harmful to humans and are effective in killing many types of bacteria. The drugs work by “deceiving” the bacteria in the following way. To survive, many micro‑organisms use p‑aminobenzoic acid to synthesize folic acid, a coenzyme in a number of biochemical processes. These micro‑organisms cannot distinguish between sulfa drugs and p‑aminobenzoic acid; so, when the drug is administered, the bacteria use it to produce a compound which has a structure similar to that of folic acid, but which is unable to act as a coenzyme in essential biochemical processes. The result is that many of the bacteria’s amino acids and nucleotides cannot be made, and the bacteria die. Amino acids are discussed in Chapter 26; nucleotides are discussed in Section 28.1.
An “arenediazonium salt” is formed by the reaction of an aromatic amine with nitrous acid at 0–5°C, and has the structure shown below.
Alkanediazonium salts are very unstable; therefore, arenediazonium salts are often simply referred to as diazonium salts.
As is mentioned in the textbook, arenediazonium salts are very useful intermediates from which a wide variety of aromatic compounds can be prepared. You should be thoroughly familiar with the use of diazonium salts to prepare each of the classes of compounds. In addition, you should be aware that fluoroarenes can also be prepared from diazonium salts, as follows:
In this case the diazonium salt is prepared using fluoroboric acid, HBF4, and sodium nitrite. The thermal decomposition of the salt, called the Schiemann reaction, can be quite hazardous.
Note that the IUPAC‑preferred name for cuprous chloride is copper(I) chloride; similarly, cuprous cyanide is called copper(I) cyanide.
Hypophosphorous acid has the structure shown below:
The presence of the letters “azo” in a compound’s name usually implies that a nitrogen‑nitrogen double bond is present in its structure. Azo compounds t in which two aryl groups are joined by an $\ce{-}$N$\ce{=}$N$\ce{-}$ linkage are usually very colourful.
Overreaction of Aniline
Arylamines are very reactive towards electrophilic aromomatic substitution. The strongest activating and ortho/para-directing substituents are the amino (-NH2) and hydroxyl (-OH) groups. Direct nitration of phenol (hydroxybenzene) by dilute nitric acid gives modest yields of nitrated phenols and considerable oxidative decomposition to tarry materials; aniline (aminobenzene) is largely destroyed. Monobromination of both phenol and aniline is difficult to control, with di- and tri-bromo products forming readily.
Because of their high nucleophilic reactivity, aniline and phenol undergo substitution reactions with iodine, a halogen that is normally unreactive with benzene derivatives. The mixed halogen iodine chloride (ICl) provides a more electrophilic iodine moiety, and is effective in iodinating aromatic rings having less powerful activating substituents.
C6H5–NH2 + I2 + NaHCO3 p-I–C6H4–NH2 + NaI + CO2 + H2O
In addition to overreactivity, we have previously seen (Section 16.3) that Friedel-Crafts reactions employing AlCl3 catalyst does not work with aniline. A salt complex forms and prevents electrophilic aromatic substitution.
Both this problem and the aniline overreactivity can be circumvented by first going through the corresponding amide.
Modifying the Influence of Strong Activating Groups
By acetylating the heteroatom substituent on aniline, its activating influence can be substantially attenuated. For example, acetylation of aniline gives acetanilide (first step in the following equation), which undergoes nitration at low temperature, yielding the para-nitro product in high yield. The modifying acetyl group can then be removed by acid-catalyzed hydrolysis (last step), to yield para-nitroaniline. Although the activating influence of the amino group has been reduced by this procedure, the acetyl derivative remains an ortho/para-directing and activating substituent.
C6H5–NH2 + (CH3CO)2O
pyridine (a base)
C6H5–NHCOCH3
HNO3 , 5 ºC
p-O2N–C6H4–NHCOCH3
H3O(+) & heat
p-O2N–C6H4–NH
The following diagram illustrates how the acetyl group acts to attenuate the overall electron donating character of oxygen and nitrogen. The non-bonding valence electron pairs that are responsible for the high reactivity of these compounds (blue arrows) are diverted to the adjacent carbonyl group (green arrows). However, the overall influence of the modified substituent is still activating and ortho/para-directing.
.
Sulfa Drug Synthesis
Sulfa drugs are an important group of synthetic antimicrobial agents (pharmaceuticals) that contain the sulfonamide group. The synthesis of sulfanilamide (a sulfa drug) illustrates how the reactivity of aniline can be modified to make possible an electrophilic aromatic substitution. The corresponding acetanilide undergoes chlorosulfonation. The resulting 4-acetamidobenzenesulfanyl chloride is treated with ammonia to replace the chlorine with an amino group and affords 4-acetamidobenzenesulfonamide. The subsequent hydrolysis of the sulfonamide produces the sulfanilamide.
Diazonium Salts: The Sandmeyer Reaction
Aryl diazonium salts are important intermediates. They are prepared in cold (0 º to 10 ºC) aqueous solution, and generally react with nucleophiles with loss of nitrogen. Some of the more commonly used substitution reactions are shown in the following diagram. Since the leaving group (N2) is thermodynamically very stable, these reactions are energetically favored. Those substitution reactions that are catalyzed by cuprous salts are known as Sandmeyer reactions. Fluoride substitution occurs on treatment with BF4(–), a reaction known as the Schiemann reaction. Stable diazonium tetrafluoroborate salts may be isolated, and on heating these lose nitrogen to give an arylfluoride product. The top reaction with hypophosphorus acid, H3PO2, is noteworthy because it achieves the reductive removal of an amino (or nitro) group. Unlike the nucleophilic substitution reactions, this reduction probably proceeds by a radical mechanism.
These aryl diazonium substitution reactions significantly expand the tactics available for the synthesis of polysubstituted benzene derivatives. Consider the following options:
1. The usual precursor to an aryl amine is the corresponding nitro compound. A nitro substituent deactivates an aromatic ring and directs electrophilic substitution to meta locations.
2. Reduction of a nitro group to an amine may be achieved in several ways. The resulting amine substituent strongly activates an aromatic ring and directs electrophilic substitution to ortho & para locations.
3. The activating character of an amine substituent may be attenuated by formation of an amide derivative (reversible), or even changed to deactivating and meta-directing by formation of a quaternary-ammonium salt (irreversible).
4. Conversion of an aryl amine to a diazonium ion intermediate allows it to be replaced by a variety of different groups (including hydrogen), which may in turn be used in subsequent reactions.
The following examples illustrate some combined applications of these options to specific cases. You should try to conceive a plausible reaction sequence for each. Once you have done so, you may check suggested answers below.
Answer 1:
It should be clear that the methyl substituent will eventually be oxidized to a carboxylic acid function. The timing is important, since a methyl substituent is ortho/para-directing and the carboxyl substituent is meta-directing. The cyano group will be introduced by a diazonium intermediate, so a nitration followed by reduction to an amine must precede this step.
Answer 2:
The hydroxyl group is a strong activating substituent and would direct aromatic ring chlorination to locations ortho & para to itself, leading to the wrong product. As an alternative, the nitro group is not only meta-directing, it can be converted to a hydroxyl group by way of a diazonium intermediate. The resulting strategy is self evident.
Answer 3:
Selective introduction of a fluorine is best achieved by treating a diazonium intermediate with boron tetrafluoride anion. To get the necessary intermediate we need to make p-nitroaniline. Since the nitro substituent on the starting material would direct a new substituent to a meta-location, we must first reduce it to an ortho/para-directing amino group. Amino groups are powerful activating substituents, so we deactivate it by acetylation before nitration. The acetyl substituent also protects the initial amine function from reaction with nitrous acid later on. It is removed in the last step.
Answer 4:
Polybromination of benzene would lead to ortho/para substitution. In order to achieve the mutual meta-relationship of three bromines, it is necessary to introduce a powerful ortho/para-directing prior to bromination, and then remove it following the tribromination. An amino group is ideal for this purpose. Reductive removal of the diazonium group may be accomplished in several ways (three are shown).
Answer 5:
The propyl substituent is best introduced by Friedel-Crafts acylation followed by reduction, and this cannot be carried out in the presence of a nitro substituent. Since an acyl substituent is a meta-director, it is logical to use this property to locate the nitro and chloro groups before reducing the carbonyl moiety. The same reduction method can be used to reduce both the nitro group (to an amine) and the carbonyl group to propyl. We have already seen the use of diazonium intermediates as precursors to phenols.
Answer 6:
Aromatic iodination can only be accomplished directly on highly activated benzene compounds, such as aniline, or indirectly by way of a diazonium intermediate. Once again, a deactivated amino group is the precursor of p-nitroaniline (prb.#3). This aniline derivative requires the more electrophilic iodine chloride (ICl) for ortho-iodination because of the presence of a deactivating nitro substituent. Finally, the third iodine is introduced by the diazonium ion procedure.
Diazonium Coupling Reactions
A resonance description of diazonium ions shows that the positive charge is delocalized over the two nitrogen atoms. It is not possible for nucleophiles to bond to the inner nitrogen, but bonding (or coupling) of negative nucleophiles to the terminal nitrogen gives neutral azo compounds. As shown in the following equation, this coupling to the terminal nitrogen should be relatively fast and reversible. The azo products may exist as E / Z stereoisomers. In practice it is found that the E-isomer predominates at equilibrium.
Unless these azo products are trapped or stabilized in some manner, reversal to the diazonium ion and slow nucleophilic substitution at carbon (with irreversible nitrogen loss) will be the ultimate course of reaction, as described in the previous section. For example, if phenyldiazonium bisufate is added rapidly to a cold solution of sodium hydroxide a relatively stable solution of sodium phenyldiazoate (the conjugate base of the initially formed diazoic acid) is obtained. Lowering the pH of this solution regenerates phenyldiazoic acid (pKa ca. 7), which disassociates back to the diazonium ion and eventually undergoes substitution, generating phenol.
C6H5N2(+) HSO4(–) + NaOH (cold solution) C6H5N2–OH + NaOH (cold) C6H5N2–O(–) Na(+)
phenyldiazonium bisulfate phenyldiazoic acid sodium phenyldiazoate
Aryl diazonium salts may be reduced to the corresponding hydrazines by mild reducing agents such as sodium bisulfite, stannous chloride or zinc dust. The bisulfite reduction may proceed by an initial sulfur-nitrogen coupling, as shown in the following equation.
Ar-N2(+) X(–)
NaHSO3
Ar-N=N-SO3H
NaHSO3
Ar-NH-NH-SO3H
H2O
Ar-NH-NH2 + H2SO4
The most important application of diazo coupling reactions is electrophilic aromatic substitution of activated benzene derivatives by diazonium electrophiles. The products of such reactions are highly colored aromatic azo compounds that find use as synthetic dyestuffs, commonly referred to as azo dyes. Azobenzene (Y=Z=H) is light orange; however, the color of other azo compounds may range from red to deep blue depending on the nature of the aromatic rings and the substituents they carry. Azo compounds may exist as cis/trans isomer pairs, but most of the well-characterized and stable compounds are trans.
Some examples of azo coupling reactions are shown below. A few simple rules are helpful in predicting the course of such reactions:
1. At acid pH (< 6) an amino group is a stronger activating substituent than a hydroxyl group (i.e. a phenol). At alkaline pH (> 7.5) phenolic functions are stronger activators, due to increased phenoxide base concentration.
2. Coupling to an activated benzene ring occurs preferentially para to the activating group if that location is free. Otherwise ortho-coupling will occur.
3. Naphthalene normally undergoes electrophilic substitution at an alpha-location more rapidly than at beta-sites; however, ortho-coupling is preferred. See the diagram for examples of α / β notation in naphthalenes.
You should try to conceive a plausible product structure for each of the following couplings. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/22%3A_Chemistry_of_the_Benzene_Substituents%3A_Alkylbenzenes_Phenols_and_Benzenamines/22.11%3A_%09Electrophilic_Substitution_with_Arenediazonium_Salts%3A_Di.txt |
Ester Enolates and the Claisen Condensation: Synthesis of β-Dicarbonyl Compounds; Acyl Anion Equivalents
23: Ester Enolates and the Claisen Condensation
Because esters can contain α hydrogens they can undergo a condensation reaction similar to the aldol reaction called a Claisen Condensation. In a fashion similar to the aldol, one ester acts as a nucleophile while a second ester acts as the electrophile. During the reaction a new carbon-carbon bond is formed. The product is a β-keto ester. A major difference with the aldol reaction is the fact that hydroxide cannot be used as a base because it could possibly react with the ester. Instead, an alkoxide version of the alcohol used to synthesize the ester is used to prevent transesterification side products.
Claisen Condensation
Basic reaction
Going from reactants to products simply
Example 1: Claisen Condensation
Claisen Condensation Mechanism
1) Enolate formation
2) Nucleophilic attack
3) Removal of leaving group
Crossed Claisen Condensation
Claisen condensations between different ester reactants are called Crossed Claisen reactions. Crossed Claisen reactions in which both reactants can serve as donors and acceptors generally give complex mixtures. Because of this most Crossed Claisen reactions are usually not performed unless one reactant has no alpha hydrogens.
Example 3: Crossed Claisen Condensation
Dieckmann Condensation
A diester can undergo an intramolecular reaction called a Dieckmann condensation.
Example 2: Dieckman Condensation
23.2: B-Dicarbonyl Compounds as Synthetic Intermediates
Enolates can act as a nucleophile in SN2 type reactions. Overall an α hydrogen is replaced with an alkyl group. This reaction is one of the more important for enolates because a carbon-carbon bond is formed. These alkylations are affected by the same limitations as SN2 reactions previously discussed. A good leaving group, Chloride, Bromide, Iodide, Tosylate, should be used. Also, secondary and tertiary leaving groups should not be used because of poor reactivity and possible competition with elimination reactions. Lastly, it is important to use a strong base, such as LDA or sodium amide, for this reaction. Using a weaker base such as hydroxide or an alkoxide leaves the possibility of multiple alkylation’s occurring.
Example 1: Alpha Alkylation
Mechanism
1) Enolate formation
2) Sn2 attack
Alkylation of Unsymmetrical Ketones
Unsymmetrical ketones can be regioselctively alkylated to form one major product depending on the reagents.
Treatment with LDA in THF at -78oC tends to form the less substituted kinetic enolate.
Using sodium ethoxide in ethanol at room temperature forms the more substituted thermodynamic enolate.
Problems
1) Please write the structure of the product for the following reactions.
Answers
1)
Contributors
Malonic ester is a reagent specifically used in a reaction which converts alkyl halides to carboxylic acids called the Malonic Ester Synthesis. Malonic ester synthesis is a synthetic procedure used to convert a compound that has the general structural formula 1 into a carboxylic acid that has the general structural formula 2.
• R1 = alkyl group
• L = leaving group
The group —CH2CO2H in 2 is contributed by a malonic ester, hence the term malonic ester synthesis.
• R2 = alkyl, aryl
Mechanism
Malonic ester synthesis consists of four consecutive reactions that can be carried out in the same pot.
• reaction 1: acid-base reaction
• reaction 2: nucleophilic substitution
• reaction 3: ester hydrolysis (using saponification)
• reaction 4: decarboxylation
eg:
reaction 1:
reaction 2:
reaction 3:
reaction 4:
A more direct method to convert 3 into 4 is the reaction of 3 with the enolate ion (5) of ethyl acetate followed by hydrolysis of the resultant ester.
However, the generation of 5 from ethyl acetate quantitatively in high yield is not an easy task because the reaction requires a very strong base, such as LDA, and must be carried out at very low temperature under strictly anhydrous conditions.
Malonic ester synthesis provides a more convenient alternative to convert 3 to 4.
Malonic ester synthesis can be adapted to synthesize compounds that have the general structural formula 6.
R3, R4 = identical or different alkyl groups
eg:
reaction 1:
reaction 2:
reaction 1 (repeat):
reaction 2 (repeat):
reaction 3:
reaction 4:
Malonic Ester Synthesis
Due to the fact that Malonic ester’s α hydrogens are adjacent to two carbonyls, they can be deprotonated by sodium ethoxide (NaOEt) to form Sodio Malonic Ester.
Because Sodio Malonic Ester is an enolate, it can then be alkylated with alkyl halides.
After alkylation the product can be converted to a dicarboxylic acid through saponification and subsequently one of the carboxylic acids can be removed through a decarboxylation step.
Mechanism
1) Saponification
2) Decarboxylation
3) Tautomerization
All of the steps together form the Malonic ester synthesis.
$RX \rightarrow RCH_2CO_2H$
Example
• Bernard E. Hoogenboom , Phillip J. Ihrig , Arne N. Langsjoen , Carol J. Linn and Stephen D. Mulder, The malonic ester synthesis in the undergraduate laboratory, J. Chem. Educ., 1991, 68 (8), p 689
Contributors
The acetoacetic ester synthesis allows for the conversion of ethyl acetoacetate into a methyl ketone with one or two alkyl groups on the alpha carbon.
Steps
1) Deprotonation with ethoxide
2) Alkylation via and SN2 Reaction
3) Hydrolysis and decarboxylation
Addition of a second alky group
After the first step and additional alkyl group can be added prior to the decarboxylation step. Overall this allows for the addition of two different alkyl groups.
23.3: B-Dicarbonyl Anion Chemistry: Michael Additions
Basic reaction of 1,4 addition
In 1,4 addition the Nucleophile is added to the carbon β to the carbonyl while the hydrogen is added to the carbon α to the carbonyl.
Mechanism for 1,4 addition
1) Nucleophilic attack on the carbon β to the carbonyl
2) Proton Transfer
Here we can see why this addition is called 1,4. The nucleophile bonds to the carbon in the one position and the hydrogen adds to the oxygen in the four position.
3) Tautomerization
Enolates undergo 1,4 addition to α, β-unsaturated carbonyl compounds is a process called a Michael addition. The reaction is named after American chemist Arthur Michael (1853-1942).
Contributors
Many times the product of a Michael addition produces a dicarbonyl which can then undergo an intramolecular aldol reaction. These two processes together in one reaction creates two new carbon-carbon bonds and also creates a ring. Ring-forming reactions are called annulations after the Latin work for ring annulus. The reaction is named after English chemist Sir Robert Robinson (1886-1975) who developed it. He received the Nobel prize in chemistry in 1947. Remember that during annulations five and six membered rings are preferred. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/23%3A_Ester_Enolates_and_the_Claisen_Condensation/23.1%3A_B-Dicarbonyl__Compounds%3A_Claisen__Condensations.txt |
Solutions to exercises
Thiamine diphosphate (TPP) is another very important coenzyme which, like PLP, acts as an electron sink to stabilize key carbanion intermediates. The important part of the TPP molecule from a catalytic standpoint is its thiazole ring.
14.5A: The benzoin condensation reaction
In a very simple experiment that can performed in an undergraduate organic chemistry lab, benzaldehyde (a liquid compound at room temperature that is used as an artificial cherry flavoring) self-condenses to form benzoin (a crystalline solid) when stirred in ethanol with a catalytic amount of sodium hydroxide and thiamine.
The laboratory synthesis of benzoin is interesting because it mimics the mechanism of TPP-dependant enzymatic reactions in biological systems. This reaction is not catalyzed by an enzyme – rather, the thiamine molecule acts on its own, playing a similar catalytic role to that played by its diphosphate ester cousin (TPP) in enzymatic reactions.
Look carefully at the connectivity of the starting compounds and product in the benzoin condensation. Essentially what has happened is that the carbonyl carbon of one benzaldehyde molecule has somehow been turned into a nucleophile, and has attacked a second benzaldehyde molecule in a nucleophilic carbonyl addition reaction.
This probably seems quite strange - we know that carbonyl carbons are good electrophiles, but how can they be nucleophilic? For the aldehyde carbon to be a nucleophile, it would have to be deprotonated, and become a carbonyl anion. But aldehyde protons are not at all acidic!
The thiamine catalyst is the key: it allows the formation of what is essentially the equivalent of a nucleophilic benzaldehyde carbanion. Let's follow the benzoin condensation reaction mechanism through step-by-step, and see how thiamine accomplishes this task.
The important part of the thiamine molecule is the thiazole ring (look again at the structure of thiamine diphosphate on the previous page), thus we will draw thiamine (and later, thiamine diphosphate) using R groups to depict the unreactive parts of the molecule. The first step of the benzoin condensation is deprotonation of thiamine by hydroxide.
It may surprise you to learn that this proton is acidic. The reason for its acidity lies partly in the ability of the neighboring sulfur atom to accept, in its open d orbitals, some of the excess electron density of the conjugate base. Another reason is that the positive charge on the nitrogen helps to stabilize the negative charge on the conjugate base. The deprotonated thiazole is called an ylide, which is a species with adjacent positively and negatively charged atoms.
The negatively charged carbon on the thiazole ylide next attacks the carbonyl of the first benzaldehyde molecule in a nucleophilic carbonyl addition (from here on out, thiamine will be colored green in order to help you to focus on the chemistry going on with the substrate).
In the resulting molecule, what used to be the aldehyde hydrogen is now acidic - this is so because, when the proton is abstracted by hydroxide, the negative charge on the conjugate base is stabilized by resonance with the positively-charged thiazole ring of thiamine. This is the function of thiamine: it acts as an electron sink, accepting electron density so as to allow for the formation of what amounts to a carbonyl anion.
Now the first benzaldehyde molecule, assisted by thiamine, can finally act as a nucleophile, attacking the carbonyl of a second benzaldehyde (step 3).
Once this is accomplished, the thiamine ylide can be kicked off as the original carbonyl on the first benzaldehyde re-forms (step 4). The thiamine ylide is now free to catalyze another reaction.
14.5B: The transketolase reaction
Now let's look at a biochemical reaction carried out by an enzyme called transketolase, with the assistance of thiamine diphosphate. Transketolase is one of a series of enzymes (along with ribulose-5-phosphate-3-epimerase, which we considered in section 13.2B) in the 'Calvin cycle' of carbon fixation in plants. Animals and bacteria also use transketolase in sugar metabolism. In the transketolase reaction, a 2-carbon unit (in the smaller, red box) is transferred between two sugar molecules:
First, let's consider how this reaction might hypothetically proceed without the assistance of the TPP cofactor. The breaking off of the 2-carbon unit from fructose-6-phosphate might be depicted as a kind of retro-aldol event:
This would lead to glyceraldehyde-3-phosphate, the first product, plus an intermediate in the form of a carbonyl anion.
Continuing to think hypothetically, this strange 2-carbon ionic intermediate could attack the glyceraldehyde-3-phosphate carbonyl, resulting in sedoheptulose-7-phosphate, the second product.
Everything in this hypothetical scenario is fine, chemically speaking, with one major exception: the carbonyl anion intermediate. It would be an extremely high energy, unlikely species. But we have seen something like this before, haven't we, in the non-enzymatic benzoin condensation reaction above! This is exactly the kind of carbanion that thiamine (this time in its diphosphate form, TPP) makes possible. The carbonyl anion is generated in different ways in the two reactions: in the benzoin condensation it is the result of the deprotonation of benzaldehyde, while in the transketolase it results from a retro-aldol-like cleavage. But that doesn't really matter - what does matter is that in each case, a thiazole ring is present to modify the starting carbonyl group and act as an electron sink, allowing it to take on a negative charge. Here, then, is the real (as opposed to hypothetical) transketolase reaction, with the role of TPP revealed.
Make sure that you can follow the electron movement throughout the mechanism, that you can see how TPP acts as an electron sink cofactor, and that you clearly recognize the mechanistic parallels to the benzoin condensation.
14.5C: Pyruvate decarboxylase
The thiamine diphosphate coenzyme also assists in the decarboxylation of an acyl group, such as in this reaction catalyzed by pyruvate decarboxylase (this is a key reaction in the fermentation of glucose to ethanol by yeast):
In this example, the TPP-stabilized carbonyl carbanion simply acts as a base rather than as a nucleophile, abstracting a proton from an enzymatic acid to form acetaldehyde.
Example 14.7
Propose a mechanism for the reaction catalyzed by pyruvate decarboxylase.
14.5D: Synthetic parallel - carbonyl nucleophiles via dithiane anions
Nature uses thiamine to generate the equivalent of a nucleophilic carbonyl anion, but with the specific exception of the benzoin condensation, a chemist working in an organic synthesis laboratory before the mid-1970's had no equivalent procedure. In 1975, E. J. Corey and Dieter Seebach reported that they had developed a method to accomplish reactions such as the following, in which aldehyde carbons act as nucleophiles in SN2 and carbonyl addition reactions (J. Org. Chem. 1975, 40, 231).
In this technique, the aldehyde is first converted to a cyclic thioketal using 1,3-propanedithiol and acid catalyst - this is the same as the method used to 'protect' aldehydes as cyclic acetals, discussed in section 11.4B, except that a dithiol is used instead of a diol.
The original aldehyde proton is now somewhat acidic, because the empty d orbitals on the two adjacent sulfur atoms are able to delocalize excess electron density of the conjugate base. A strong base, such as an organolithium compound (section 13.6B) will deprotonate the cyclic thioacetal, which can then act as a nucleophile, attacking an alkyl halide or a carbonyl electrophile (the latter case is illustrated below):
The thioacetal can then be hydrolyzed back to an aldehyde group, a process that is facilitated by the use of methyl iodide.
Example 14.8
Show a mechanism for the hydrolysis of a cyclic thioacetal, in the presence of catalytic acid and methyl iodide. Propose a role for methyl iodide in this reaction.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/23%3A_Ester_Enolates_and_the_Claisen_Condensation/23.4%3A_Alkanoyl__%28Acyl%29__Anion__Equivalents%3A__Preparation_of_a-Hydroxyketones.txt |
Objectives
After completing this section, you should be able to
1. classify a specific carbohydrate as being a monosaccharide, disaccharide, trisaccharide, etc., given the structure of the carbohydrate or sufficient information about its structure.
2. classify a monosaccharide according to the number of carbon atoms present and whether it contains an aldehyde or ketone group.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• aldose
• disaccharide
• ketose
• monosaccharide (simple sugar)
• polysaccharide
Carbohydrates are the most abundant class of organic compounds found in living organisms. They originate as products of photosynthesis, an endothermic reductive condensation of carbon dioxide requiring light energy and the pigment chlorophyll.
$nCO_2 + n H_2O + Energy \rightarrow C_nH_{2n}O_n + nO_2$
As noted here, the formulas of many carbohydrates can be written as carbon hydrates, $C_n(H_2O)_n$, hence their name. The carbohydrates are a major source of metabolic energy, both for plants and for animals that depend on plants for food. Aside from the sugars and starches that meet this vital nutritional role, carbohydrates also serve as a structural material (cellulose), a component of the energy transport compound ATP/ADP, recognition sites on cell surfaces, and one of three essential components of DNA and RNA.
Carbohydrates are called saccharides or, if they are relatively small, sugars. Several classifications of carbohydrates have proven useful, and are outlined in the following table.
Complexity
Simple Carbohydrates
monosaccharides
Complex Carbohydrates
disaccharides, oligosaccharides
& polysaccharides
Size
Tetrose
C4 sugars
Pentose
C5 sugars
Hexose
C6 sugars
Heptose
C7 sugars
etc.
C=O Function
Aldose
sugars having an aldehyde function or an acetal equivalent.
Ketose
sugars having a ketone function or an acetal equivalent.
Reactivity
Reducing
sugars oxidized by Tollens' reagent (or Benedict's or Fehling's reagents).
Non-reducing
sugars not oxidized by Tollens' or other reagents.
Contributors and Attributions
Objectives
After completing this section, you should be able to
1. draw the Fischer projection of a monosaccharide, given its wedge‑and‑broken‑line structure or a molecular model.
2. draw the wedge‑and‑broken‑line structure of a monosaccharide, given its Fischer projection or a molecular model.
3. construct a molecular model of a monosaccharide, given its Fischer projection or wedge‑and‑broken‑line structure.
Key Terms
Make certain that you can define, and use in context, the key term below.
• Fischer projection
Study Notes
When studying this section, use your molecular model set to assist you in visualizing the structures of the compounds that are discussed. It is important that you be able to determine whether two apparently different Fischer projections represent two different structures or one single structure. Often the simplest way to check is to construct a molecular model corresponding to each projection formula, and then compare the two models.
The problem of drawing three-dimensional configurations on a two-dimensional surface, such as a piece of paper, has been a long-standing concern of chemists. The wedge and hatched line notations we have been using are effective, but can be troublesome when applied to compounds having many chiral centers. As part of his Nobel Prize-winning research on carbohydrates, the great German chemist Emil Fischer, devised a simple notation that is still widely used. In a Fischer projection drawing, the four bonds to a chiral carbon make a cross with the carbon atom at the intersection of the horizontal and vertical lines. The two horizontal bonds are directed toward the viewer (forward of the stereogenic carbon). The two vertical bonds are directed behind the central carbon (away from the viewer). Since this is not the usual way in which we have viewed such structures, the following diagram shows how a stereogenic carbon positioned in the common two-bonds-in-a-plane orientation ( x–C–y define the reference plane ) is rotated into the Fischer projection orientation (the far right formula). When writing Fischer projection formulas it is important to remember these conventions. Since the vertical bonds extend away from the viewer and the horizontal bonds toward the viewer, a Fischer structure may only be turned by 180º within the plane, thus maintaining this relationship. The structure must not be flipped over or rotated by 90º.
In the above diagram, if x = CO2H, y = CH3, a = H & b = OH, the resulting formula describes (R)-(–)-lactic acid. The mirror-image formula, where x = CO2H, y = CH3, a = OH & b = H, would, of course, represent (S)-(+)-lactic acid.
The Fischer Projection consists of both horizontal and vertical lines, where the horizontal lines represent the atoms that are pointed toward the viewer while the vertical line represents atoms that are pointed away from the viewer. The point of intersection between the horizontal and vertical lines represents the central carbon.
Using the Fischer projection notation, the stereoisomers of 2-methylamino-1-phenylpropanol are drawn in the following manner. Note that it is customary to set the longest carbon chain as the vertical bond assembly.
The usefulness of this notation to Fischer, in his carbohydrate studies, is evident in the following diagram. There are eight stereoisomers of 2,3,4,5-tetrahydroxypentanal, a group of compounds referred to as the aldopentoses. Since there are three chiral centers in this constitution, we should expect a maximum of 23 stereoisomers. These eight stereoisomers consist of four sets of enantiomers. If the configuration at C-4 is kept constant (R in the examples shown here), the four stereoisomers that result will be diastereomers. Fischer formulas for these isomers, which Fischer designated as the "D"-family, are shown in the diagram. Each of these compounds has an enantiomer, which is a member of the "L"-family so, as expected, there are eight stereoisomers in all. Determining whether a chiral carbon is R or S may seem difficult when using Fischer projections, but it is actually quite simple. If the lowest priority group (often a hydrogen) is on a vertical bond, the configuration is given directly from the relative positions of the three higher-ranked substituents. If the lowest priority group is on a horizontal bond, the positions of the remaining groups give the wrong answer (you are in looking at the configuration from the wrong side), so you simply reverse it.
The aldopentose structures drawn above are all diastereomers. A more selective term, epimer, is used to designate diastereomers that differ in configuration at only one chiral center. Thus, ribose and arabinose are epimers at C-2, and arabinose and lyxose are epimers at C-3. However, arabinose and xylose are not epimers, since their configurations differ at both C-2 and C-3.
How to make Fischer Projections
To make a Fischer Projection, it is easier to show through examples than through words. Lets start with the first example, turning a 3D structure of ethane into a 2D Fischer Projection.
Example 25.2.1
Start by mentally converting a 3D structure into a Dashed-Wedged Line Structure. Remember, the atoms that are pointed toward the viewer would be designated with a wedged lines and the ones pointed away from the viewer are designated with dashed lines.
Figure A Figure B
Notice the red balls (atoms) in Figure A above are pointed away from the screen. These atoms will be designated with dashed lines like those in Figure B by number 2 and 6. The green balls (atoms) are pointed toward the screen. These atoms will be designated with wedged lines like those in Figure B by number 3 and 5. The blue atoms are in the plane of the screen so they are designated with straight lines.
Now that we have our Dashed- Wedged Line Structure, we can convert it to a Fischer Projection. However, before we can convert this Dashed-Wedged Line Structure into a Fischer Projection, we must first convert it to a “flat” Dashed-Wedged Line Structure. Then from there we can draw our Fischer Projection. Lets start with a more simpler example. Instead of using the ethane shown in Figure A and B, we will start with a methane. The reason being is that it allows us to only focus on one central carbon, which make things a little bit easier.
Figure C Figure D
Lets start with this 3D image and work our way to a dashed-wedged image. Start by imagining yourself looking directly at the central carbon from the left side as shown in Figure C. It should look something like Figure D. Now take this Figure D and flatten it out on the surface of the paper and you should get an image of a cross.
As a reminder, the horizontal line represents atoms that are coming out of the paper and the vertical line represents atoms that are going into the paper. The cross image to the right of the arrow is a Fischer projection.
Contributors and Attributions
Objectives
After completing this section, you should be able to
1. identify a specific enantioner of a monosaccharide as being D or L, given its Fischer projection.
2. identify the limitations of the D, L system of nomenclature for carbohydrates.
3. assign an R or S configuration to each of the chiral carbon atoms present in a monosaccharide, given its Fischer projection.
4. draw the Fischer projection formula for a monosaccharide, given its systematic name, complete with the configuration of each chiral carbon atom.
5. construct a molecular model of a monosaccharide, given its systematic name, complete with the configuration of each chiral carbon atom.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• D sugar
• L sugar
Study Notes
If you find that you have forgotten the meanings of terms such as dextrorotatory and polarimeter, refer back to Section 5.3 in which the fundamentals of optical activity were introduced.
How would you set about the task of deciding whether each chiral carbon has an R or an S configuration? True, you could use molecular models, but suppose that a model set had not been available—what would you have done then?
One approach is to focus on the carbon atom of interest and sketch a three-dimensional representation of the configuration around that atom, remembering the convention used in Fischer projections: vertical lines represent bonds going into the page, and horizontal lines represent bonds coming out of the page. Thus, the configuration around carbon atom 2 in structure a can be represented as follows:
In your mind, you should be able to imagine how this molecule would look if it was rotated so that the bonds that are shown as coming out of the page are now in the plane of the page. [One possible way of doing this is to try and imagine how the molecule would look if it was viewed from a point at the bottom of the page.] What you should see in your mind is a representation similar to the one drawn below.
To determine whether the configuration about the central carbon atom is R or S, we must rotate the molecule so that the group with the lowest priority (H), is directed away from the viewer. This effect can be achieved by keeping the hydroxyl group in its present position and moving each of the other three groups one position clockwise.
The Cahn-Ingold-Prelog order of priority for the three remaining groups is OH > CHO > CH(OH)CH2OH; thus, we see that we could trace out a counterclockwise path going from the highest-priority group to the second- and third-highest, and we conclude that the central carbon atom has an S configuration.
The Configuration of Glucose
The four chiral centers in glucose indicate there may be as many as sixteen (24) stereoisomers having this constitution. These would exist as eight diastereomeric pairs of enantiomers, and the initial challenge was to determine which of the eight corresponded to glucose. This challenge was accepted and met in 1891 by the German chemist Emil Fischer. His successful negotiation of the stereochemical maze presented by the aldohexoses was a logical tour de force, and it is fitting that he received the 1902 Nobel Prize for chemistry for this accomplishment. One of the first tasks faced by Fischer was to devise a method of representing the configuration of each chiral center in an unambiguous manner. To this end, he invented a simple technique for drawing chains of chiral centers, that we now call the Fischer projection formula. Click on this link for a review.
At the time Fischer undertook the glucose project it was not possible to establish the absolute configuration of an enantiomer. Consequently, Fischer made an arbitrary choice for (+)-glucose and established a network of related aldose configurations that he called the D-family. The mirror images of these configurations were then designated the L-family of aldoses. To illustrate using present day knowledge, Fischer projection formulas and names for the D-aldose family (three to six-carbon atoms) are shown below, with the asymmetric carbon atoms (chiral centers) colored red. The last chiral center in an aldose chain (farthest from the aldehyde group) was chosen by Fischer as the D / L designator site. If the hydroxyl group in the projection formula pointed to the right, it was defined as a member of the D-family. A left directed hydroxyl group (the mirror image) then represented the L-family. Fischer's initial assignment of the D-configuration had a 50:50 chance of being right, but all his subsequent conclusions concerning the relative configurations of various aldoses were soundly based. In 1951 x-ray fluorescence studies of (+)-tartaric acid, carried out in the Netherlands by Johannes Martin Bijvoet, proved that Fischer's choice was correct.
It is important to recognize that the sign of a compound's specific rotation (an experimental number) does not correlate with its configuration (D or L). It is a simple matter to measure an optical rotation with a polarimeter. Determining an absolute configuration usually requires chemical interconversion with known compounds by stereospecific reaction paths.
Contributors and Attributions
Objectives
After completing this section, you should be able to
1. draw the structures of all possible aldotetroses, aldopentoses, and aldohexoses, without necessarily being able to assign names to the individual compounds.
2. draw the Fischer projection of D‑glyceraldehyde, D‑ribose and D‑glucose from memory.
The four chiral centers in glucose indicate there may be as many as sixteen (24) stereoisomers having this constitution. These would exist as eight diastereomeric pairs of enantiomers, and the initial challenge was to determine which of the eight corresponded to glucose. This challenge was accepted and met in 1891 by the German chemist Emil Fischer. His successful negotiation of the stereochemical maze presented by the aldohexoses was a logical tour de force, and it is fitting that he received the 1902 Nobel Prize for chemistry for this accomplishment. One of the first tasks faced by Fischer was to devise a method of representing the configuration of each chiral center in an unambiguous manner. To this end, he invented a simple technique for drawing chains of chiral centers, that we now call the Fischer projection formula.
At the time Fischer undertook the glucose project it was not possible to establish the absolute configuration of an enantiomer. Consequently, Fischer made an arbitrary choice for (+)-glucose and established a network of related aldose configurations that he called the D-family. The mirror images of these configurations were then designated the L-family of aldoses. To illustrate using present day knowledge, Fischer projection formulas and names for the D-aldose family (three to six-carbon atoms) are shown below, with the asymmetric carbon atoms (chiral centers) colored red.
The last chiral center in an aldose chain (farthest from the aldehyde group) was chosen by Fischer as the D / L designator site. If the hydroxyl group in the projection formula pointed to the right, it was defined as a member of the D-family. A left directed hydroxyl group (the mirror image) then represented the L-family. Fischer's initial assignment of the D-configuration had a 50:50 chance of being right, but all his subsequent conclusions concerning the relative configurations of various aldoses were soundly based. In 1951 x-ray fluorescence studies of (+)-tartaric acid, carried out in the Netherlands by Johannes Martin Bijvoet (pronounced "buy foot"), proved that Fischer's choice was correct.
It is important to recognize that the sign of a compound's specific rotation (an experimental number) does not correlate with its configuration (D or L). It is a simple matter to measure an optical rotation with a polarimeter. Determining an absolute configuration usually requires chemical interconversion with known compounds by stereospecific reaction paths. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/24%3A_Carbohydrates%3A_Polyfunctional_Compounds_in_Nature/24.01%3A_Names__and__Structures_of_Carbohydrates.txt |
Objectives
After completing this section, you should be able to
1. determine whether a given monosaccharide will exist as a pyranose or furanose.
2. draw the cyclic pyranose form of a monosaccharide, given its Fischer projection.
3. draw the Fischer projection of a monosaccharide, given its cyclic pyranose form.
4. draw, from memory, the cyclic pyranose form of D‑glucose.
5. determine whether a given cyclic pyranose form represents the D or L form of the monosaccharide concerned.
6. describe the phenomenon known as mutarotation.
7. explain, through the use of chemical equations, exactly what happens at the molecular level during the mutarotation process.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• alpha anomer
• anomer
• anomeric centre
• beta anomer
• furanose
• mutarotation
• pyranose
Study Notes
If necessary, before you attempt to study this section, review the formation of hemiacetals discussed in Section 19.10.
As noted above, the preferred structural form of many monosaccharides may be that of a cyclic hemiacetal. Five and six-membered rings are favored over other ring sizes because of their low angle and eclipsing strain. Cyclic structures of this kind are termed furanose (five-membered) or pyranose (six-membered), reflecting the ring size relationship to the common heterocyclic compounds furan and pyran shown on the right. Ribose, an important aldopentose, commonly adopts a furanose structure, as shown in the following illustration. By convention for the D-family, the five-membered furanose ring is drawn in an edgewise projection with the ring oxygen positioned away from the viewer. The anomeric carbon atom (colored red here) is placed on the right. The upper bond to this carbon is defined as beta, the lower bond then is alpha.
The cyclic pyranose forms of various monosaccharides are often drawn in a flat projection known as a Haworth formula, after the British chemist, Norman Haworth. As with the furanose ring, the anomeric carbon is placed on the right with the ring oxygen to the back of the edgewise view. In the D-family, the alpha and beta bonds have the same orientation defined for the furanose ring (beta is up & alpha is down). These Haworth formulas are convenient for displaying stereochemical relationships, but do not represent the true shape of the molecules. We know that these molecules are actually puckered in a fashion we call a chair conformation. Examples of four typical pyranose structures are shown below, both as Haworth projections and as the more representative chair conformers. The anomeric carbons are colored red.
The size of the cyclic hemiacetal ring adopted by a given sugar is not constant, but may vary with substituents and other structural features. Aldolhexoses usually form pyranose rings and their pentose homologs tend to prefer the furanose form, but there are many counter examples. The formation of acetal derivatives illustrates how subtle changes may alter this selectivity. A pyranose structure for D-glucose is drawn in the rose-shaded box on the left. Acetal derivatives have been prepared by acid-catalyzed reactions with benzaldehyde and acetone. As a rule, benzaldehyde forms six-membered cyclic acetals, whereas acetone prefers to form five-membered acetals. The top equation shows the formation and some reactions of the 4,6-O-benzylidene acetal, a commonly employed protective group. A methyl glycoside derivative of this compound (see below) leaves the C-2 and C-3 hydroxyl groups exposed to reactions such as the periodic acid cleavage, shown as the last step. The formation of an isopropylidene acetal at C-1 and C-2, center structure, leaves the C-3 hydroxyl as the only unprotected function. Selective oxidation to a ketone is then possible. Finally, direct di-O-isopropylidene derivatization of glucose by reaction with excess acetone results in a change to a furanose structure in which the C-3 hydroxyl is again unprotected. However, the same reaction with D-galactose, shown in the blue-shaded box, produces a pyranose product in which the C-6 hydroxyl is unprotected. Both derivatives do not react with Tollens' reagent. This difference in behavior is attributed to the cis-orientation of the C-3 and C-4 hydroxyl groups in galactose, which permits formation of a less strained five-membered cyclic acetal, compared with the trans-C-3 and C-4 hydroxyl groups in glucose. Derivatizations of this kind permit selective reactions to be conducted at different locations in these highly functionalized molecules.
Anomers of Simple Sugars: Mutarotation of Glucose
Figure 1: Cyclization of D-Glucose. D-Glucose can be represented with a Fischer projection (a) or three dimensionally (b). By reacting the OH group on the fifth carbon atom with the aldehyde group, the cyclic monosaccharide (c) is produced.
When a straight-chain monosaccharide, such as any of the structures shown in Figure 1, forms a cyclic structure, the carbonyl oxygen atom may be pushed either up or down, giving rise to two stereoisomers, as shown in Figure 2. The structure shown on the left side of Figure 2, with the OH group on the first carbon atom projected downward, represent what is called the alpha (α) form. The structures on the right side, with the OH group on the first carbon atom pointed upward, is the beta (β) form. These two stereoisomers of a cyclic monosaccharide are known as anomers; they differ in structure around the anomeric carbon—that is, the carbon atom that was the carbonyl carbon atom in the straight-chain form.
Figure 2: Monosaccharides. In an aqueous solution, monosaccharides exist as an equilibrium mixture of three forms. The interconversion between the forms is known as mutarotation, which is shown for D-glucose (a) and D-fructose (b).
It is possible to obtain a sample of crystalline glucose in which all the molecules have the α structure or all have the β structure. The α form melts at 146°C and has a specific rotation of +112°, while the β form melts at 150°C and has a specific rotation of +18.7°. When the sample is dissolved in water, however, a mixture is soon produced containing both anomers as well as the straight-chain form, in dynamic equilibrium (part (a) of Figure 2). You can start with a pure crystalline sample of glucose consisting entirely of either anomer, but as soon as the molecules dissolve in water, they open to form the carbonyl group and then reclose to form either the α or the β anomer. The opening and closing repeats continuously in an ongoing interconversion between anomeric forms and is referred to as mutarotation (Latin mutare, meaning “to change”). At equilibrium, the mixture consists of about 36% α-D-glucose, 64% β-D-glucose, and less than 0.02% of the open-chain aldehyde form. The observed rotation of this solution is +52.7°.
Even though only a small percentage of the molecules are in the open-chain aldehyde form at any time, the solution will nevertheless exhibit the characteristic reactions of an aldehyde. As the small amount of free aldehyde is used up in a reaction, there is a shift in the equilibrium to yield more aldehyde. Thus, all the molecules may eventually react, even though very little free aldehyde is present at a time.
Commonly, (e.g., in Figures 1 and 2) the cyclic forms of sugars are depicted using a convention first suggested by Walter N. Haworth, an English chemist. The molecules are drawn as planar hexagons with a darkened edge representing the side facing toward the viewer. The structure is simplified to show only the functional groups attached to the carbon atoms. Any group written to the right in a Fischer projection appears below the plane of the ring in a Haworth projection, and any group written to the left in a Fischer projection appears above the plane in a Haworth projection.
The difference between the α and the β forms of sugars may seem trivial, but such structural differences are often crucial in biochemical reactions. This explains why we can get energy from the starch in potatoes and other plants but not from cellulose, even though both starch and cellulose are polysaccharides composed of glucose molecules linked together.
Summary
Monosaccharides that contain five or more carbons atoms form cyclic structures in aqueous solution. Two cyclic stereoisomers can form from each straight-chain monosaccharide; these are known as anomers. In an aqueous solution, an equilibrium mixture forms between the two anomers and the straight-chain structure of a monosaccharide in a process known as mutarotation.
Exercises
1. Draw the cyclic structure for β-D-glucose. Identify the anomeric carbon.
2. Given that the aldohexose D-mannose differs from D-glucose only in the configuration at the second carbon atom, draw the cyclic structure for α-D-mannose.
2. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/24%3A_Carbohydrates%3A_Polyfunctional_Compounds_in_Nature/24.02%3A_Conformations_and__Cyclic__Forms__of_Sugars.txt |
Learning Objectives
• Define what is meant by anomers and describe how they are formed.
• Explain what is meant by mutarotation.
So far we have represented monosaccharides as linear molecules, but many of them also adopt cyclic structures. This conversion occurs because of the ability of aldehydes and ketones to react with alcohols:
You might wonder why the aldehyde reacts with the OH group on the fifth carbon atom rather than the OH group on the second carbon atom next to it. Recall that cyclic alkanes containing five or six carbon atoms in the ring are the most stable. The same is true for monosaccharides that form cyclic structures: rings consisting of five or six carbon atoms are the most stable.
When a straight-chain monosaccharide, such as any of the structures shown in Figure \(1\), forms a cyclic structure, the carbonyl oxygen atom may be pushed either up or down, giving rise to two stereoisomers, as shown in Figure \(2\). The structure shown on the left side of Figure \(2\), with the OH group on the first carbon atom projected downward, represent what is called the alpha (α) form. The structures on the right side, with the OH group on the first carbon atom pointed upward, is the beta (β) form. These two stereoisomers of a cyclic monosaccharide are known as anomers; they differ in structure around the anomeric carbon—that is, the carbon atom that was the carbonyl carbon atom in the straight-chain form.
It is possible to obtain a sample of crystalline glucose in which all the molecules have the α structure or all have the β structure. The α form melts at 146°C and has a specific rotation of +112°, while the β form melts at 150°C and has a specific rotation of +18.7°. When the sample is dissolved in water, however, a mixture is soon produced containing both anomers as well as the straight-chain form, in dynamic equilibrium (part (a) of Figure \(2\)). You can start with a pure crystalline sample of glucose consisting entirely of either anomer, but as soon as the molecules dissolve in water, they open to form the carbonyl group and then reclose to form either the α or the β anomer. The opening and closing repeats continuously in an ongoing interconversion between anomeric forms and is referred to as mutarotation (Latin mutare, meaning “to change”). At equilibrium, the mixture consists of about 36% α-D-glucose, 64% β-D-glucose, and less than 0.02% of the open-chain aldehyde form. The observed rotation of this solution is +52.7°.
Even though only a small percentage of the molecules are in the open-chain aldehyde form at any time, the solution will nevertheless exhibit the characteristic reactions of an aldehyde. As the small amount of free aldehyde is used up in a reaction, there is a shift in the equilibrium to yield more aldehyde. Thus, all the molecules may eventually react, even though very little free aldehyde is present at a time.
Commonly, (e.g., in Figures \(1\) and \(2\)) the cyclic forms of sugars are depicted using a convention first suggested by Walter N. Haworth, an English chemist. The molecules are drawn as planar hexagons with a darkened edge representing the side facing toward the viewer. The structure is simplified to show only the functional groups attached to the carbon atoms. Any group written to the right in a Fischer projection appears below the plane of the ring in a Haworth projection, and any group written to the left in a Fischer projection appears above the plane in a Haworth projection.
The difference between the α and the β forms of sugars may seem trivial, but such structural differences are often crucial in biochemical reactions. This explains why we can get energy from the starch in potatoes and other plants but not from cellulose, even though both starch and cellulose are polysaccharides composed of glucose molecules linked together.
Summary
Monosaccharides that contain five or more carbons atoms form cyclic structures in aqueous solution. Two cyclic stereoisomers can form from each straight-chain monosaccharide; these are known as anomers. In an aqueous solution, an equilibrium mixture forms between the two anomers and the straight-chain structure of a monosaccharide in a process known as mutarotation. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/24%3A_Carbohydrates%3A_Polyfunctional_Compounds_in_Nature/24.03%3A_Anomers__of_Simple__Sugars%3A__Mutarotation_of_Glucose.txt |
Objectives
After completing this section, you should be able to
1. write equations to illustrate that the hydroxyl groups of carbohydrates can react to form esters and ethers.
2. identify the product formed when a given monosaccharide is reacted with acetic anhydride or with silver oxide and an alkyl halide.
3. identify the reagents required to convert a given monosaccharide to its ester or ether.
1. write an equation to show how a monosaccharide can be converted to a glycoside using an alcohol and an acid catalyst.
2. identify the product formed when a given monosaccharide is treated with an alcohol and an acid catalyst.
3. write a detailed mechanism for the formation of a glycoside by the reaction of the cyclic form of a monosaccharide with an alcohol and an acid catalyst.
1. identify the ester formed by phosphorylation in biologically important compounds.
1. identify the product formed when a given monosaccharide is reduced with sodium borohydride.
2. identify the monosaccharide which should be reduced in order to form a given polyalcohol (alditol).
1. explain that a sugar with an aldehyde or hemiacetal can be oxidized to the corresponding carboxylic acid (also known as aldonic acid). Note: The sugar is able to reduce an oxidizing agent, and is thus called a reducing sugar. Tests for reducing sugars include the use of Tollens’ reagent, Fehling’s reagent and Benedict’s reagent.
2. explain why certain ketoses, such as fructose, behave as reducing sugars even though they do not contain an aldehyde group.
3. identify warm HNO3 as the reagent needed to form dicarboxylic acid (an aldaric acid).
1. describe the chain‑lengthening effect of the Kiliani‑Fischer synthesis.
2. predict the product that would be produced by the Kiliani‑Fischer synthesis of a given aldose.
3. identify the aldose that would yield a given product following Kiliani‑Fischer synthesis.
1. describe the chain‑shortening effect of the Wohl degradation.
2. predict the product that would be produced by the Wohl degradation of a given aldose.
3. identify the aldose or aldoses that would yield a given product following Wohl degradation.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• aldaric acid
• aldonic acid
• alditol
• aldonic acid
• glycoside
• Kiliani-Fischer synthesis
• neighbouring group effect
• reducing sugar
• Wohl degradation
Study Notes
While several reactions are covered in this section, keep in mind that you have encountered them in previous sections. The active functional groups on monosaccharides are essentially carbonyls and hydroxyls. Although they now are a part of much larger molecules, their chemistry should be familiar.
The formation of esters and ethers is quite straightforward and should not require further clarification.
Note that glycosides are in fact acetals, and that glycoside formation is therefore analogous to acetal formation. To refresh your memory about the chemistry of acetals, quickly review Section 19.10
Ester and Ether Formation
The -OH groups on a monosaccharide can be readily converted to esters and ethers. Esterfication can be done with an acid chloride (Section 21.4) or acid anhydride (Section 21.5), while treatment with an alkyl halide by a Williamson ether synthesis (Section 18.2) leads to the ether.
Glycoside Formation
Acetal derivatives formed when a monosaccharide reacts with an alcohol in the presence of an acid catalyst are called glycosides. This reaction is illustrated for glucose and methanol in the diagram below. In naming of glycosides, the "ose" suffix of the sugar name is replaced by "oside", and the alcohol group name is placed first. As is generally true for most acetals, glycoside formation involves the loss of an equivalent of water. The diether product is stable to base and alkaline oxidants such as Tollen's reagent. Since acid-catalyzed aldolization is reversible, glycosides may be hydrolyzed back to their alcohol and sugar components by aqueous acid.
The anomeric methyl glucosides are formed in an equilibrium ratio of 66% alpha to 34% beta. From the structures in the previous diagram, we see that pyranose rings prefer chair conformations in which the largest number of substituents are equatorial. In the case of glucose, the substituents on the beta-anomer are all equatorial, whereas the C-1 substituent in the alpha-anomer changes to axial. Since substituents on cyclohexane rings prefer an equatorial location over axial (methoxycyclohexane is 75% equatorial), the preference for alpha-glycopyranoside formation is unexpected, and is referred to as the anomeric effect.
Glycosides abound in biological systems. By attaching a sugar moiety to a lipid or benzenoid structure, the solubility and other properties of the compound may be changed substantially. Because of the important modifying influence of such derivatization, numerous enzyme systems, known as glycosidases, have evolved for the attachment and removal of sugars from alcohols, phenols and amines. Chemists refer to the sugar component of natural glycosides as the glycon and the alcohol component as the aglycon.
Two examples of naturally occurring glycosides and one example of an amino derivative are displayed above. Salicin, one of the oldest herbal remedies known, was the model for the synthetic analgesic aspirin. A large class of hydroxylated, aromatic oxonium cations called anthocyanins provide the red, purple and blue colors of many flowers, fruits and some vegetables. Peonin is one example of this class of natural pigments, which exhibit a pronounced pH color dependence. The oxonium moiety is only stable in acidic environments, and the color changes or disappears when base is added. The complex changes that occur when wine is fermented and stored are in part associated with glycosides of anthocyanins. Finally, amino derivatives of ribose, such as cytidine play important roles in biological phosphorylating agents, coenzymes and information transport and storage materials.
Biological Ester Formation: Phosphorylation
Recall that almost all biomolecules are charged species, which 1) keeps them water soluble, and 2) prevents them from diffusing across lipid bilayer membranes. Although many biomolecules are ionized by virtue of negatively charged carboxylate and positively charged amino groups, the most common ionic group in biologically important organic compounds is phosphate - thus the phosphorylation of alcohol groups is a critical metabolic step. In alcohol phosphorylations, ATP is almost always the phosphate donor, and the mechanism is very consistent: the alcohol oxygen acts as a nucleophile, attacking the gamma-phosphorus of ATP and expelling ADP (look again, for example, at the glucose kinase reaction that we first saw in section 10.1D).
Oxidation
As noted above, sugars may be classified as reducing or non-reducing based on their reactivity with Tollens', Benedict's or Fehling's reagents. If a sugar is oxidized by these reagents it is called reducing, since the oxidant (Ag(+) or Cu(+2)) is reduced in the reaction, as evidenced by formation of a silver mirror or precipitation of cuprous oxide. The Tollens' test is commonly used to detect aldehyde functions; and because of the facile interconversion of ketoses and aldoses under the basic conditions of this test, ketoses such as fructose also react and are classified as reducing sugars.
When the aldehyde function of an aldose is oxidized to a carboxylic acid the product is called an aldonic acid. Because of the 2º hydroxyl functions that are also present in these compounds, a mild oxidizing agent such as hypobromite must be used for this conversion (equation 1). If both ends of an aldose chain are oxidized to carboxylic acids the product is called an aldaric acid. By converting an aldose to its corresponding aldaric acid derivative, the ends of the chain become identical (this could also be accomplished by reducing the aldehyde to CH2OH, as noted below). Such an operation will disclose any latent symmetry in the remaining molecule. Thus, ribose, xylose, allose and galactose yield achiral aldaric acids which are, of course, not optically active. The ribose oxidation is shown in equation 2 below.
1.
2.
3.
Other aldose sugars may give identical chiral aldaric acid products, implying a unique configurational relationship. The examples of arabinose and lyxose shown in equation 3 above illustrate this result. Remember, a Fischer projection formula may be rotated by 180º in the plane of projection without changing its configuration.
Reduction
Sodium borohydride reduction of an aldose makes the ends of the resulting alditol chain identical, HOCH2(CHOH)nCH2OH, thereby accomplishing the same configurational change produced by oxidation to an aldaric acid. Thus, allitol and galactitol from reduction of allose and galactose are achiral, and altrose and talose are reduced to the same chiral alditol. A summary of these redox reactions, and derivative nomenclature is given in the following table.
Table: Derivatives of \(HOCH_2(CHOH)_nCHO\)
\(HOBr\) Oxidation
\(\longrightarrow\)
\(HOCH_2(CHOH)_nCO_2H\)
an Aldonic Acid
\(HNO_3\) Oxidation
\(\longrightarrow\)
\(H_2OC(CHOH)_nCO_2H\)
an Aldaric Acid
\(NaBH_4\) Reduction
\(\longrightarrow\)
\(HOCH_2(CHOH)_nCH_2OH\)
an Alditol
Chain Shortening and Lengthening
1. Ruff Degradation
2. Kiliani-Fischer Synthesis
These two procedures permit an aldose of a given size to be related to homologous smaller and larger aldoses. The importance of these relationships may be seen in the array of aldose structures presented earlier, where the structural connections are given by the dashed blue lines. Thus Ruff degradation of the pentose arabinose gives the tetrose erythrose. Working in the opposite direction, a Kiliani-Fischer synthesis applied to arabinose gives a mixture of glucose and mannose. An alternative chain shortening procedure known as the Wohl degradation is essentially the reverse of the Kiliani-Fischer synthesis.
Note that in the Kiliani-Fischer synthesis the first step is to generate a cyanohydrin intermediate, which is then has its nitrile group hydrolyzed to the carboxylic acid. From there the cyclic ester (lactone) is formed and reduced to the final products (epimers).
Wohl Degradation
The ability to shorten (degrade) an aldose chain by one carbon was an important tool in the structure elucidation of carbohydrates. This was commonly accomplished by the Ruff procedure. An interesting alternative technique, known as the Wohl degradation has also been used. The following equation illustrates the application of this procedure to the aldopentose, arabinose. Based on your knowledge of carbonyl chemistry, and considering that the Wohl degradation is in essence the reverse of the Kiliani-Fischer synthesis. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/24%3A_Carbohydrates%3A_Polyfunctional_Compounds_in_Nature/24.04%3A_Polyfunctional__Chemistry__of_Sugars%3A__Oxidation__to_Carboxylic__Acids.txt |
Objectives
After completing this section, you should be able to
1. write equations to illustrate that the hydroxyl groups of carbohydrates can react to form esters and ethers.
2. identify the product formed when a given monosaccharide is reacted with acetic anhydride or with silver oxide and an alkyl halide.
3. identify the reagents required to convert a given monosaccharide to its ester or ether.
1. write an equation to show how a monosaccharide can be converted to a glycoside using an alcohol and an acid catalyst.
2. identify the product formed when a given monosaccharide is treated with an alcohol and an acid catalyst.
3. write a detailed mechanism for the formation of a glycoside by the reaction of the cyclic form of a monosaccharide with an alcohol and an acid catalyst.
1. identify the ester formed by phosphorylation in biologically important compounds.
1. identify the product formed when a given monosaccharide is reduced with sodium borohydride.
2. identify the monosaccharide which should be reduced in order to form a given polyalcohol (alditol).
1. explain that a sugar with an aldehyde or hemiacetal can be oxidized to the corresponding carboxylic acid (also known as aldonic acid). Note: The sugar is able to reduce an oxidizing agent, and is thus called a reducing sugar. Tests for reducing sugars include the use of Tollens’ reagent, Fehling’s reagent and Benedict’s reagent.
2. explain why certain ketoses, such as fructose, behave as reducing sugars even though they do not contain an aldehyde group.
3. identify warm HNO3 as the reagent needed to form dicarboxylic acid (an aldaric acid).
1. describe the chain‑lengthening effect of the Kiliani‑Fischer synthesis.
2. predict the product that would be produced by the Kiliani‑Fischer synthesis of a given aldose.
3. identify the aldose that would yield a given product following Kiliani‑Fischer synthesis.
1. describe the chain‑shortening effect of the Wohl degradation.
2. predict the product that would be produced by the Wohl degradation of a given aldose.
3. identify the aldose or aldoses that would yield a given product following Wohl degradation.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• aldaric acid
• aldonic acid
• alditol
• aldonic acid
• glycoside
• Kiliani-Fischer synthesis
• neighbouring group effect
• reducing sugar
• Wohl degradation
Study Notes
While several reactions are covered in this section, keep in mind that you have encountered them in previous sections. The active functional groups on monosaccharides are essentially carbonyls and hydroxyls. Although they now are a part of much larger molecules, their chemistry should be familiar.
The formation of esters and ethers is quite straightforward and should not require further clarification.
Note that glycosides are in fact acetals, and that glycoside formation is therefore analogous to acetal formation. To refresh your memory about the chemistry of acetals, quickly review Section 19.10
Ester and Ether Formation
The -OH groups on a monosaccharide can be readily converted to esters and ethers. Esterfication can be done with an acid chloride (Section 21.4) or acid anhydride (Section 21.5), while treatment with an alkyl halide by a Williamson ether synthesis (Section 18.2) leads to the ether.
Glycoside Formation
Acetal derivatives formed when a monosaccharide reacts with an alcohol in the presence of an acid catalyst are called glycosides. This reaction is illustrated for glucose and methanol in the diagram below. In naming of glycosides, the "ose" suffix of the sugar name is replaced by "oside", and the alcohol group name is placed first. As is generally true for most acetals, glycoside formation involves the loss of an equivalent of water. The diether product is stable to base and alkaline oxidants such as Tollen's reagent. Since acid-catalyzed aldolization is reversible, glycosides may be hydrolyzed back to their alcohol and sugar components by aqueous acid.
The anomeric methyl glucosides are formed in an equilibrium ratio of 66% alpha to 34% beta. From the structures in the previous diagram, we see that pyranose rings prefer chair conformations in which the largest number of substituents are equatorial. In the case of glucose, the substituents on the beta-anomer are all equatorial, whereas the C-1 substituent in the alpha-anomer changes to axial. Since substituents on cyclohexane rings prefer an equatorial location over axial (methoxycyclohexane is 75% equatorial), the preference for alpha-glycopyranoside formation is unexpected, and is referred to as the anomeric effect.
Glycosides abound in biological systems. By attaching a sugar moiety to a lipid or benzenoid structure, the solubility and other properties of the compound may be changed substantially. Because of the important modifying influence of such derivatization, numerous enzyme systems, known as glycosidases, have evolved for the attachment and removal of sugars from alcohols, phenols and amines. Chemists refer to the sugar component of natural glycosides as the glycon and the alcohol component as the aglycon.
Two examples of naturally occurring glycosides and one example of an amino derivative are displayed above. Salicin, one of the oldest herbal remedies known, was the model for the synthetic analgesic aspirin. A large class of hydroxylated, aromatic oxonium cations called anthocyanins provide the red, purple and blue colors of many flowers, fruits and some vegetables. Peonin is one example of this class of natural pigments, which exhibit a pronounced pH color dependence. The oxonium moiety is only stable in acidic environments, and the color changes or disappears when base is added. The complex changes that occur when wine is fermented and stored are in part associated with glycosides of anthocyanins. Finally, amino derivatives of ribose, such as cytidine play important roles in biological phosphorylating agents, coenzymes and information transport and storage materials.
Biological Ester Formation: Phosphorylation
Recall that almost all biomolecules are charged species, which 1) keeps them water soluble, and 2) prevents them from diffusing across lipid bilayer membranes. Although many biomolecules are ionized by virtue of negatively charged carboxylate and positively charged amino groups, the most common ionic group in biologically important organic compounds is phosphate - thus the phosphorylation of alcohol groups is a critical metabolic step. In alcohol phosphorylations, ATP is almost always the phosphate donor, and the mechanism is very consistent: the alcohol oxygen acts as a nucleophile, attacking the gamma-phosphorus of ATP and expelling ADP (look again, for example, at the glucose kinase reaction that we first saw in section 10.1D).
Oxidation
As noted above, sugars may be classified as reducing or non-reducing based on their reactivity with Tollens', Benedict's or Fehling's reagents. If a sugar is oxidized by these reagents it is called reducing, since the oxidant (Ag(+) or Cu(+2)) is reduced in the reaction, as evidenced by formation of a silver mirror or precipitation of cuprous oxide. The Tollens' test is commonly used to detect aldehyde functions; and because of the facile interconversion of ketoses and aldoses under the basic conditions of this test, ketoses such as fructose also react and are classified as reducing sugars.
When the aldehyde function of an aldose is oxidized to a carboxylic acid the product is called an aldonic acid. Because of the 2º hydroxyl functions that are also present in these compounds, a mild oxidizing agent such as hypobromite must be used for this conversion (equation 1). If both ends of an aldose chain are oxidized to carboxylic acids the product is called an aldaric acid. By converting an aldose to its corresponding aldaric acid derivative, the ends of the chain become identical (this could also be accomplished by reducing the aldehyde to CH2OH, as noted below). Such an operation will disclose any latent symmetry in the remaining molecule. Thus, ribose, xylose, allose and galactose yield achiral aldaric acids which are, of course, not optically active. The ribose oxidation is shown in equation 2 below.
1.
2.
3.
Other aldose sugars may give identical chiral aldaric acid products, implying a unique configurational relationship. The examples of arabinose and lyxose shown in equation 3 above illustrate this result. Remember, a Fischer projection formula may be rotated by 180º in the plane of projection without changing its configuration.
Reduction
Sodium borohydride reduction of an aldose makes the ends of the resulting alditol chain identical, HOCH2(CHOH)nCH2OH, thereby accomplishing the same configurational change produced by oxidation to an aldaric acid. Thus, allitol and galactitol from reduction of allose and galactose are achiral, and altrose and talose are reduced to the same chiral alditol. A summary of these redox reactions, and derivative nomenclature is given in the following table.
Table: Derivatives of \(HOCH_2(CHOH)_nCHO\)
\(HOBr\) Oxidation
\(\longrightarrow\)
\(HOCH_2(CHOH)_nCO_2H\)
an Aldonic Acid
\(HNO_3\) Oxidation
\(\longrightarrow\)
\(H_2OC(CHOH)_nCO_2H\)
an Aldaric Acid
\(NaBH_4\) Reduction
\(\longrightarrow\)
\(HOCH_2(CHOH)_nCH_2OH\)
an Alditol
Chain Shortening and Lengthening
1. Ruff Degradation
2. Kiliani-Fischer Synthesis
These two procedures permit an aldose of a given size to be related to homologous smaller and larger aldoses. The importance of these relationships may be seen in the array of aldose structures presented earlier, where the structural connections are given by the dashed blue lines. Thus Ruff degradation of the pentose arabinose gives the tetrose erythrose. Working in the opposite direction, a Kiliani-Fischer synthesis applied to arabinose gives a mixture of glucose and mannose. An alternative chain shortening procedure known as the Wohl degradation is essentially the reverse of the Kiliani-Fischer synthesis.
Note that in the Kiliani-Fischer synthesis the first step is to generate a cyanohydrin intermediate, which is then has its nitrile group hydrolyzed to the carboxylic acid. From there the cyclic ester (lactone) is formed and reduced to the final products (epimers).
Wohl Degradation
The ability to shorten (degrade) an aldose chain by one carbon was an important tool in the structure elucidation of carbohydrates. This was commonly accomplished by the Ruff procedure. An interesting alternative technique, known as the Wohl degradation has also been used. The following equation illustrates the application of this procedure to the aldopentose, arabinose. Based on your knowledge of carbonyl chemistry, and considering that the Wohl degradation is in essence the reverse of the Kiliani-Fischer synthesis. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/24%3A_Carbohydrates%3A_Polyfunctional_Compounds_in_Nature/24.06%3A_Reduction__of_Monosaccharides__to_Alditols.txt |
Objectives
After completing this section, you should be able to
1. write equations to illustrate that the hydroxyl groups of carbohydrates can react to form esters and ethers.
2. identify the product formed when a given monosaccharide is reacted with acetic anhydride or with silver oxide and an alkyl halide.
3. identify the reagents required to convert a given monosaccharide to its ester or ether.
1. write an equation to show how a monosaccharide can be converted to a glycoside using an alcohol and an acid catalyst.
2. identify the product formed when a given monosaccharide is treated with an alcohol and an acid catalyst.
3. write a detailed mechanism for the formation of a glycoside by the reaction of the cyclic form of a monosaccharide with an alcohol and an acid catalyst.
1. identify the ester formed by phosphorylation in biologically important compounds.
1. identify the product formed when a given monosaccharide is reduced with sodium borohydride.
2. identify the monosaccharide which should be reduced in order to form a given polyalcohol (alditol).
1. explain that a sugar with an aldehyde or hemiacetal can be oxidized to the corresponding carboxylic acid (also known as aldonic acid). Note: The sugar is able to reduce an oxidizing agent, and is thus called a reducing sugar. Tests for reducing sugars include the use of Tollens’ reagent, Fehling’s reagent and Benedict’s reagent.
2. explain why certain ketoses, such as fructose, behave as reducing sugars even though they do not contain an aldehyde group.
3. identify warm HNO3 as the reagent needed to form dicarboxylic acid (an aldaric acid).
1. describe the chain‑lengthening effect of the Kiliani‑Fischer synthesis.
2. predict the product that would be produced by the Kiliani‑Fischer synthesis of a given aldose.
3. identify the aldose that would yield a given product following Kiliani‑Fischer synthesis.
1. describe the chain‑shortening effect of the Wohl degradation.
2. predict the product that would be produced by the Wohl degradation of a given aldose.
3. identify the aldose or aldoses that would yield a given product following Wohl degradation.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• aldaric acid
• aldonic acid
• alditol
• aldonic acid
• glycoside
• Kiliani-Fischer synthesis
• neighbouring group effect
• reducing sugar
• Wohl degradation
Study Notes
While several reactions are covered in this section, keep in mind that you have encountered them in previous sections. The active functional groups on monosaccharides are essentially carbonyls and hydroxyls. Although they now are a part of much larger molecules, their chemistry should be familiar.
The formation of esters and ethers is quite straightforward and should not require further clarification.
Note that glycosides are in fact acetals, and that glycoside formation is therefore analogous to acetal formation. To refresh your memory about the chemistry of acetals, quickly review Section 19.10
Ester and Ether Formation
The -OH groups on a monosaccharide can be readily converted to esters and ethers. Esterfication can be done with an acid chloride (Section 21.4) or acid anhydride (Section 21.5), while treatment with an alkyl halide by a Williamson ether synthesis (Section 18.2) leads to the ether.
Glycoside Formation
Acetal derivatives formed when a monosaccharide reacts with an alcohol in the presence of an acid catalyst are called glycosides. This reaction is illustrated for glucose and methanol in the diagram below. In naming of glycosides, the "ose" suffix of the sugar name is replaced by "oside", and the alcohol group name is placed first. As is generally true for most acetals, glycoside formation involves the loss of an equivalent of water. The diether product is stable to base and alkaline oxidants such as Tollen's reagent. Since acid-catalyzed aldolization is reversible, glycosides may be hydrolyzed back to their alcohol and sugar components by aqueous acid.
The anomeric methyl glucosides are formed in an equilibrium ratio of 66% alpha to 34% beta. From the structures in the previous diagram, we see that pyranose rings prefer chair conformations in which the largest number of substituents are equatorial. In the case of glucose, the substituents on the beta-anomer are all equatorial, whereas the C-1 substituent in the alpha-anomer changes to axial. Since substituents on cyclohexane rings prefer an equatorial location over axial (methoxycyclohexane is 75% equatorial), the preference for alpha-glycopyranoside formation is unexpected, and is referred to as the anomeric effect.
Glycosides abound in biological systems. By attaching a sugar moiety to a lipid or benzenoid structure, the solubility and other properties of the compound may be changed substantially. Because of the important modifying influence of such derivatization, numerous enzyme systems, known as glycosidases, have evolved for the attachment and removal of sugars from alcohols, phenols and amines. Chemists refer to the sugar component of natural glycosides as the glycon and the alcohol component as the aglycon.
Two examples of naturally occurring glycosides and one example of an amino derivative are displayed above. Salicin, one of the oldest herbal remedies known, was the model for the synthetic analgesic aspirin. A large class of hydroxylated, aromatic oxonium cations called anthocyanins provide the red, purple and blue colors of many flowers, fruits and some vegetables. Peonin is one example of this class of natural pigments, which exhibit a pronounced pH color dependence. The oxonium moiety is only stable in acidic environments, and the color changes or disappears when base is added. The complex changes that occur when wine is fermented and stored are in part associated with glycosides of anthocyanins. Finally, amino derivatives of ribose, such as cytidine play important roles in biological phosphorylating agents, coenzymes and information transport and storage materials.
Biological Ester Formation: Phosphorylation
Recall that almost all biomolecules are charged species, which 1) keeps them water soluble, and 2) prevents them from diffusing across lipid bilayer membranes. Although many biomolecules are ionized by virtue of negatively charged carboxylate and positively charged amino groups, the most common ionic group in biologically important organic compounds is phosphate - thus the phosphorylation of alcohol groups is a critical metabolic step. In alcohol phosphorylations, ATP is almost always the phosphate donor, and the mechanism is very consistent: the alcohol oxygen acts as a nucleophile, attacking the gamma-phosphorus of ATP and expelling ADP (look again, for example, at the glucose kinase reaction that we first saw in section 10.1D).
Oxidation
As noted above, sugars may be classified as reducing or non-reducing based on their reactivity with Tollens', Benedict's or Fehling's reagents. If a sugar is oxidized by these reagents it is called reducing, since the oxidant (Ag(+) or Cu(+2)) is reduced in the reaction, as evidenced by formation of a silver mirror or precipitation of cuprous oxide. The Tollens' test is commonly used to detect aldehyde functions; and because of the facile interconversion of ketoses and aldoses under the basic conditions of this test, ketoses such as fructose also react and are classified as reducing sugars.
When the aldehyde function of an aldose is oxidized to a carboxylic acid the product is called an aldonic acid. Because of the 2º hydroxyl functions that are also present in these compounds, a mild oxidizing agent such as hypobromite must be used for this conversion (equation 1). If both ends of an aldose chain are oxidized to carboxylic acids the product is called an aldaric acid. By converting an aldose to its corresponding aldaric acid derivative, the ends of the chain become identical (this could also be accomplished by reducing the aldehyde to CH2OH, as noted below). Such an operation will disclose any latent symmetry in the remaining molecule. Thus, ribose, xylose, allose and galactose yield achiral aldaric acids which are, of course, not optically active. The ribose oxidation is shown in equation 2 below.
1.
2.
3.
Other aldose sugars may give identical chiral aldaric acid products, implying a unique configurational relationship. The examples of arabinose and lyxose shown in equation 3 above illustrate this result. Remember, a Fischer projection formula may be rotated by 180º in the plane of projection without changing its configuration.
Reduction
Sodium borohydride reduction of an aldose makes the ends of the resulting alditol chain identical, HOCH2(CHOH)nCH2OH, thereby accomplishing the same configurational change produced by oxidation to an aldaric acid. Thus, allitol and galactitol from reduction of allose and galactose are achiral, and altrose and talose are reduced to the same chiral alditol. A summary of these redox reactions, and derivative nomenclature is given in the following table.
Table: Derivatives of \(HOCH_2(CHOH)_nCHO\)
\(HOBr\) Oxidation
\(\longrightarrow\)
\(HOCH_2(CHOH)_nCO_2H\)
an Aldonic Acid
\(HNO_3\) Oxidation
\(\longrightarrow\)
\(H_2OC(CHOH)_nCO_2H\)
an Aldaric Acid
\(NaBH_4\) Reduction
\(\longrightarrow\)
\(HOCH_2(CHOH)_nCH_2OH\)
an Alditol
Chain Shortening and Lengthening
1. Ruff Degradation
2. Kiliani-Fischer Synthesis
These two procedures permit an aldose of a given size to be related to homologous smaller and larger aldoses. The importance of these relationships may be seen in the array of aldose structures presented earlier, where the structural connections are given by the dashed blue lines. Thus Ruff degradation of the pentose arabinose gives the tetrose erythrose. Working in the opposite direction, a Kiliani-Fischer synthesis applied to arabinose gives a mixture of glucose and mannose. An alternative chain shortening procedure known as the Wohl degradation is essentially the reverse of the Kiliani-Fischer synthesis.
Note that in the Kiliani-Fischer synthesis the first step is to generate a cyanohydrin intermediate, which is then has its nitrile group hydrolyzed to the carboxylic acid. From there the cyclic ester (lactone) is formed and reduced to the final products (epimers).
Wohl Degradation
The ability to shorten (degrade) an aldose chain by one carbon was an important tool in the structure elucidation of carbohydrates. This was commonly accomplished by the Ruff procedure. An interesting alternative technique, known as the Wohl degradation has also been used. The following equation illustrates the application of this procedure to the aldopentose, arabinose. Based on your knowledge of carbonyl chemistry, and considering that the Wohl degradation is in essence the reverse of the Kiliani-Fischer synthesis. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/24%3A_Carbohydrates%3A_Polyfunctional_Compounds_in_Nature/24.07%3A_Carbonyl__Condensations__with_Amine_Derivatives.txt |
Objectives
After completing this section, you should be able to
1. write equations to illustrate that the hydroxyl groups of carbohydrates can react to form esters and ethers.
2. identify the product formed when a given monosaccharide is reacted with acetic anhydride or with silver oxide and an alkyl halide.
3. identify the reagents required to convert a given monosaccharide to its ester or ether.
1. write an equation to show how a monosaccharide can be converted to a glycoside using an alcohol and an acid catalyst.
2. identify the product formed when a given monosaccharide is treated with an alcohol and an acid catalyst.
3. write a detailed mechanism for the formation of a glycoside by the reaction of the cyclic form of a monosaccharide with an alcohol and an acid catalyst.
1. identify the ester formed by phosphorylation in biologically important compounds.
1. identify the product formed when a given monosaccharide is reduced with sodium borohydride.
2. identify the monosaccharide which should be reduced in order to form a given polyalcohol (alditol).
1. explain that a sugar with an aldehyde or hemiacetal can be oxidized to the corresponding carboxylic acid (also known as aldonic acid). Note: The sugar is able to reduce an oxidizing agent, and is thus called a reducing sugar. Tests for reducing sugars include the use of Tollens’ reagent, Fehling’s reagent and Benedict’s reagent.
2. explain why certain ketoses, such as fructose, behave as reducing sugars even though they do not contain an aldehyde group.
3. identify warm HNO3 as the reagent needed to form dicarboxylic acid (an aldaric acid).
1. describe the chain‑lengthening effect of the Kiliani‑Fischer synthesis.
2. predict the product that would be produced by the Kiliani‑Fischer synthesis of a given aldose.
3. identify the aldose that would yield a given product following Kiliani‑Fischer synthesis.
1. describe the chain‑shortening effect of the Wohl degradation.
2. predict the product that would be produced by the Wohl degradation of a given aldose.
3. identify the aldose or aldoses that would yield a given product following Wohl degradation.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• aldaric acid
• aldonic acid
• alditol
• aldonic acid
• glycoside
• Kiliani-Fischer synthesis
• neighbouring group effect
• reducing sugar
• Wohl degradation
Study Notes
While several reactions are covered in this section, keep in mind that you have encountered them in previous sections. The active functional groups on monosaccharides are essentially carbonyls and hydroxyls. Although they now are a part of much larger molecules, their chemistry should be familiar.
The formation of esters and ethers is quite straightforward and should not require further clarification.
Note that glycosides are in fact acetals, and that glycoside formation is therefore analogous to acetal formation. To refresh your memory about the chemistry of acetals, quickly review Section 19.10
Ester and Ether Formation
The -OH groups on a monosaccharide can be readily converted to esters and ethers. Esterfication can be done with an acid chloride (Section 21.4) or acid anhydride (Section 21.5), while treatment with an alkyl halide by a Williamson ether synthesis (Section 18.2) leads to the ether.
Glycoside Formation
Acetal derivatives formed when a monosaccharide reacts with an alcohol in the presence of an acid catalyst are called glycosides. This reaction is illustrated for glucose and methanol in the diagram below. In naming of glycosides, the "ose" suffix of the sugar name is replaced by "oside", and the alcohol group name is placed first. As is generally true for most acetals, glycoside formation involves the loss of an equivalent of water. The diether product is stable to base and alkaline oxidants such as Tollen's reagent. Since acid-catalyzed aldolization is reversible, glycosides may be hydrolyzed back to their alcohol and sugar components by aqueous acid.
The anomeric methyl glucosides are formed in an equilibrium ratio of 66% alpha to 34% beta. From the structures in the previous diagram, we see that pyranose rings prefer chair conformations in which the largest number of substituents are equatorial. In the case of glucose, the substituents on the beta-anomer are all equatorial, whereas the C-1 substituent in the alpha-anomer changes to axial. Since substituents on cyclohexane rings prefer an equatorial location over axial (methoxycyclohexane is 75% equatorial), the preference for alpha-glycopyranoside formation is unexpected, and is referred to as the anomeric effect.
Glycosides abound in biological systems. By attaching a sugar moiety to a lipid or benzenoid structure, the solubility and other properties of the compound may be changed substantially. Because of the important modifying influence of such derivatization, numerous enzyme systems, known as glycosidases, have evolved for the attachment and removal of sugars from alcohols, phenols and amines. Chemists refer to the sugar component of natural glycosides as the glycon and the alcohol component as the aglycon.
Two examples of naturally occurring glycosides and one example of an amino derivative are displayed above. Salicin, one of the oldest herbal remedies known, was the model for the synthetic analgesic aspirin. A large class of hydroxylated, aromatic oxonium cations called anthocyanins provide the red, purple and blue colors of many flowers, fruits and some vegetables. Peonin is one example of this class of natural pigments, which exhibit a pronounced pH color dependence. The oxonium moiety is only stable in acidic environments, and the color changes or disappears when base is added. The complex changes that occur when wine is fermented and stored are in part associated with glycosides of anthocyanins. Finally, amino derivatives of ribose, such as cytidine play important roles in biological phosphorylating agents, coenzymes and information transport and storage materials.
Biological Ester Formation: Phosphorylation
Recall that almost all biomolecules are charged species, which 1) keeps them water soluble, and 2) prevents them from diffusing across lipid bilayer membranes. Although many biomolecules are ionized by virtue of negatively charged carboxylate and positively charged amino groups, the most common ionic group in biologically important organic compounds is phosphate - thus the phosphorylation of alcohol groups is a critical metabolic step. In alcohol phosphorylations, ATP is almost always the phosphate donor, and the mechanism is very consistent: the alcohol oxygen acts as a nucleophile, attacking the gamma-phosphorus of ATP and expelling ADP (look again, for example, at the glucose kinase reaction that we first saw in section 10.1D).
Oxidation
As noted above, sugars may be classified as reducing or non-reducing based on their reactivity with Tollens', Benedict's or Fehling's reagents. If a sugar is oxidized by these reagents it is called reducing, since the oxidant (Ag(+) or Cu(+2)) is reduced in the reaction, as evidenced by formation of a silver mirror or precipitation of cuprous oxide. The Tollens' test is commonly used to detect aldehyde functions; and because of the facile interconversion of ketoses and aldoses under the basic conditions of this test, ketoses such as fructose also react and are classified as reducing sugars.
When the aldehyde function of an aldose is oxidized to a carboxylic acid the product is called an aldonic acid. Because of the 2º hydroxyl functions that are also present in these compounds, a mild oxidizing agent such as hypobromite must be used for this conversion (equation 1). If both ends of an aldose chain are oxidized to carboxylic acids the product is called an aldaric acid. By converting an aldose to its corresponding aldaric acid derivative, the ends of the chain become identical (this could also be accomplished by reducing the aldehyde to CH2OH, as noted below). Such an operation will disclose any latent symmetry in the remaining molecule. Thus, ribose, xylose, allose and galactose yield achiral aldaric acids which are, of course, not optically active. The ribose oxidation is shown in equation 2 below.
1.
2.
3.
Other aldose sugars may give identical chiral aldaric acid products, implying a unique configurational relationship. The examples of arabinose and lyxose shown in equation 3 above illustrate this result. Remember, a Fischer projection formula may be rotated by 180º in the plane of projection without changing its configuration.
Reduction
Sodium borohydride reduction of an aldose makes the ends of the resulting alditol chain identical, HOCH2(CHOH)nCH2OH, thereby accomplishing the same configurational change produced by oxidation to an aldaric acid. Thus, allitol and galactitol from reduction of allose and galactose are achiral, and altrose and talose are reduced to the same chiral alditol. A summary of these redox reactions, and derivative nomenclature is given in the following table.
Table: Derivatives of \(HOCH_2(CHOH)_nCHO\)
\(HOBr\) Oxidation
\(\longrightarrow\)
\(HOCH_2(CHOH)_nCO_2H\)
an Aldonic Acid
\(HNO_3\) Oxidation
\(\longrightarrow\)
\(H_2OC(CHOH)_nCO_2H\)
an Aldaric Acid
\(NaBH_4\) Reduction
\(\longrightarrow\)
\(HOCH_2(CHOH)_nCH_2OH\)
an Alditol
Chain Shortening and Lengthening
1. Ruff Degradation
2. Kiliani-Fischer Synthesis
These two procedures permit an aldose of a given size to be related to homologous smaller and larger aldoses. The importance of these relationships may be seen in the array of aldose structures presented earlier, where the structural connections are given by the dashed blue lines. Thus Ruff degradation of the pentose arabinose gives the tetrose erythrose. Working in the opposite direction, a Kiliani-Fischer synthesis applied to arabinose gives a mixture of glucose and mannose. An alternative chain shortening procedure known as the Wohl degradation is essentially the reverse of the Kiliani-Fischer synthesis.
Note that in the Kiliani-Fischer synthesis the first step is to generate a cyanohydrin intermediate, which is then has its nitrile group hydrolyzed to the carboxylic acid. From there the cyclic ester (lactone) is formed and reduced to the final products (epimers).
Wohl Degradation
The ability to shorten (degrade) an aldose chain by one carbon was an important tool in the structure elucidation of carbohydrates. This was commonly accomplished by the Ruff procedure. An interesting alternative technique, known as the Wohl degradation has also been used. The following equation illustrates the application of this procedure to the aldopentose, arabinose. Based on your knowledge of carbonyl chemistry, and considering that the Wohl degradation is in essence the reverse of the Kiliani-Fischer synthesis. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/24%3A_Carbohydrates%3A_Polyfunctional_Compounds_in_Nature/24.08%3A_Ester_and_Ether_Formation%3A__Glycosides.txt |
Objectives
After completing this section, you should be able to
1. write equations to illustrate that the hydroxyl groups of carbohydrates can react to form esters and ethers.
2. identify the product formed when a given monosaccharide is reacted with acetic anhydride or with silver oxide and an alkyl halide.
3. identify the reagents required to convert a given monosaccharide to its ester or ether.
1. write an equation to show how a monosaccharide can be converted to a glycoside using an alcohol and an acid catalyst.
2. identify the product formed when a given monosaccharide is treated with an alcohol and an acid catalyst.
3. write a detailed mechanism for the formation of a glycoside by the reaction of the cyclic form of a monosaccharide with an alcohol and an acid catalyst.
1. identify the ester formed by phosphorylation in biologically important compounds.
1. identify the product formed when a given monosaccharide is reduced with sodium borohydride.
2. identify the monosaccharide which should be reduced in order to form a given polyalcohol (alditol).
1. explain that a sugar with an aldehyde or hemiacetal can be oxidized to the corresponding carboxylic acid (also known as aldonic acid). Note: The sugar is able to reduce an oxidizing agent, and is thus called a reducing sugar. Tests for reducing sugars include the use of Tollens’ reagent, Fehling’s reagent and Benedict’s reagent.
2. explain why certain ketoses, such as fructose, behave as reducing sugars even though they do not contain an aldehyde group.
3. identify warm HNO3 as the reagent needed to form dicarboxylic acid (an aldaric acid).
1. describe the chain‑lengthening effect of the Kiliani‑Fischer synthesis.
2. predict the product that would be produced by the Kiliani‑Fischer synthesis of a given aldose.
3. identify the aldose that would yield a given product following Kiliani‑Fischer synthesis.
1. describe the chain‑shortening effect of the Wohl degradation.
2. predict the product that would be produced by the Wohl degradation of a given aldose.
3. identify the aldose or aldoses that would yield a given product following Wohl degradation.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• aldaric acid
• aldonic acid
• alditol
• aldonic acid
• glycoside
• Kiliani-Fischer synthesis
• neighbouring group effect
• reducing sugar
• Wohl degradation
Study Notes
While several reactions are covered in this section, keep in mind that you have encountered them in previous sections. The active functional groups on monosaccharides are essentially carbonyls and hydroxyls. Although they now are a part of much larger molecules, their chemistry should be familiar.
The formation of esters and ethers is quite straightforward and should not require further clarification.
Note that glycosides are in fact acetals, and that glycoside formation is therefore analogous to acetal formation. To refresh your memory about the chemistry of acetals, quickly review Section 19.10
Ester and Ether Formation
The -OH groups on a monosaccharide can be readily converted to esters and ethers. Esterfication can be done with an acid chloride (Section 21.4) or acid anhydride (Section 21.5), while treatment with an alkyl halide by a Williamson ether synthesis (Section 18.2) leads to the ether.
Glycoside Formation
Acetal derivatives formed when a monosaccharide reacts with an alcohol in the presence of an acid catalyst are called glycosides. This reaction is illustrated for glucose and methanol in the diagram below. In naming of glycosides, the "ose" suffix of the sugar name is replaced by "oside", and the alcohol group name is placed first. As is generally true for most acetals, glycoside formation involves the loss of an equivalent of water. The diether product is stable to base and alkaline oxidants such as Tollen's reagent. Since acid-catalyzed aldolization is reversible, glycosides may be hydrolyzed back to their alcohol and sugar components by aqueous acid.
The anomeric methyl glucosides are formed in an equilibrium ratio of 66% alpha to 34% beta. From the structures in the previous diagram, we see that pyranose rings prefer chair conformations in which the largest number of substituents are equatorial. In the case of glucose, the substituents on the beta-anomer are all equatorial, whereas the C-1 substituent in the alpha-anomer changes to axial. Since substituents on cyclohexane rings prefer an equatorial location over axial (methoxycyclohexane is 75% equatorial), the preference for alpha-glycopyranoside formation is unexpected, and is referred to as the anomeric effect.
Glycosides abound in biological systems. By attaching a sugar moiety to a lipid or benzenoid structure, the solubility and other properties of the compound may be changed substantially. Because of the important modifying influence of such derivatization, numerous enzyme systems, known as glycosidases, have evolved for the attachment and removal of sugars from alcohols, phenols and amines. Chemists refer to the sugar component of natural glycosides as the glycon and the alcohol component as the aglycon.
Two examples of naturally occurring glycosides and one example of an amino derivative are displayed above. Salicin, one of the oldest herbal remedies known, was the model for the synthetic analgesic aspirin. A large class of hydroxylated, aromatic oxonium cations called anthocyanins provide the red, purple and blue colors of many flowers, fruits and some vegetables. Peonin is one example of this class of natural pigments, which exhibit a pronounced pH color dependence. The oxonium moiety is only stable in acidic environments, and the color changes or disappears when base is added. The complex changes that occur when wine is fermented and stored are in part associated with glycosides of anthocyanins. Finally, amino derivatives of ribose, such as cytidine play important roles in biological phosphorylating agents, coenzymes and information transport and storage materials.
Biological Ester Formation: Phosphorylation
Recall that almost all biomolecules are charged species, which 1) keeps them water soluble, and 2) prevents them from diffusing across lipid bilayer membranes. Although many biomolecules are ionized by virtue of negatively charged carboxylate and positively charged amino groups, the most common ionic group in biologically important organic compounds is phosphate - thus the phosphorylation of alcohol groups is a critical metabolic step. In alcohol phosphorylations, ATP is almost always the phosphate donor, and the mechanism is very consistent: the alcohol oxygen acts as a nucleophile, attacking the gamma-phosphorus of ATP and expelling ADP (look again, for example, at the glucose kinase reaction that we first saw in section 10.1D).
Oxidation
As noted above, sugars may be classified as reducing or non-reducing based on their reactivity with Tollens', Benedict's or Fehling's reagents. If a sugar is oxidized by these reagents it is called reducing, since the oxidant (Ag(+) or Cu(+2)) is reduced in the reaction, as evidenced by formation of a silver mirror or precipitation of cuprous oxide. The Tollens' test is commonly used to detect aldehyde functions; and because of the facile interconversion of ketoses and aldoses under the basic conditions of this test, ketoses such as fructose also react and are classified as reducing sugars.
When the aldehyde function of an aldose is oxidized to a carboxylic acid the product is called an aldonic acid. Because of the 2º hydroxyl functions that are also present in these compounds, a mild oxidizing agent such as hypobromite must be used for this conversion (equation 1). If both ends of an aldose chain are oxidized to carboxylic acids the product is called an aldaric acid. By converting an aldose to its corresponding aldaric acid derivative, the ends of the chain become identical (this could also be accomplished by reducing the aldehyde to CH2OH, as noted below). Such an operation will disclose any latent symmetry in the remaining molecule. Thus, ribose, xylose, allose and galactose yield achiral aldaric acids which are, of course, not optically active. The ribose oxidation is shown in equation 2 below.
1.
2.
3.
Other aldose sugars may give identical chiral aldaric acid products, implying a unique configurational relationship. The examples of arabinose and lyxose shown in equation 3 above illustrate this result. Remember, a Fischer projection formula may be rotated by 180º in the plane of projection without changing its configuration.
Reduction
Sodium borohydride reduction of an aldose makes the ends of the resulting alditol chain identical, HOCH2(CHOH)nCH2OH, thereby accomplishing the same configurational change produced by oxidation to an aldaric acid. Thus, allitol and galactitol from reduction of allose and galactose are achiral, and altrose and talose are reduced to the same chiral alditol. A summary of these redox reactions, and derivative nomenclature is given in the following table.
Table: Derivatives of \(HOCH_2(CHOH)_nCHO\)
\(HOBr\) Oxidation
\(\longrightarrow\)
\(HOCH_2(CHOH)_nCO_2H\)
an Aldonic Acid
\(HNO_3\) Oxidation
\(\longrightarrow\)
\(H_2OC(CHOH)_nCO_2H\)
an Aldaric Acid
\(NaBH_4\) Reduction
\(\longrightarrow\)
\(HOCH_2(CHOH)_nCH_2OH\)
an Alditol
Chain Shortening and Lengthening
1. Ruff Degradation
2. Kiliani-Fischer Synthesis
These two procedures permit an aldose of a given size to be related to homologous smaller and larger aldoses. The importance of these relationships may be seen in the array of aldose structures presented earlier, where the structural connections are given by the dashed blue lines. Thus Ruff degradation of the pentose arabinose gives the tetrose erythrose. Working in the opposite direction, a Kiliani-Fischer synthesis applied to arabinose gives a mixture of glucose and mannose. An alternative chain shortening procedure known as the Wohl degradation is essentially the reverse of the Kiliani-Fischer synthesis.
Note that in the Kiliani-Fischer synthesis the first step is to generate a cyanohydrin intermediate, which is then has its nitrile group hydrolyzed to the carboxylic acid. From there the cyclic ester (lactone) is formed and reduced to the final products (epimers).
Wohl Degradation
The ability to shorten (degrade) an aldose chain by one carbon was an important tool in the structure elucidation of carbohydrates. This was commonly accomplished by the Ruff procedure. An interesting alternative technique, known as the Wohl degradation has also been used. The following equation illustrates the application of this procedure to the aldopentose, arabinose. Based on your knowledge of carbonyl chemistry, and considering that the Wohl degradation is in essence the reverse of the Kiliani-Fischer synthesis. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/24%3A_Carbohydrates%3A_Polyfunctional_Compounds_in_Nature/24.09%3A_Step-by-Step__Buildup__and_Degradation__of_Sugars.txt |
Objectives
After completing this section, you should be able to
1. identify disaccharides as compounds consisting of two monosaccharide units joined by a glycoside link between the C1 of one sugar and one of the hydroxyl groups of a second sugar.
2. identify the two monosaccharide units in a given disaccharide.
3. identify the type of glycoside link (e.g., 1,4′‑β) present in a given disaccharide structure.
4. draw the structure of a specific disaccharide, given the structure of the monosaccharide units and the type of glycoside link involved.
Note: If α‑ or β‑D‑glucose were one of the monosaccharide units, its structure would not be provided.
5. identify the structural feature that determines whether or not a given disaccharide behaves as a reducing sugar and undergoes mutarotation, and write equations to illustrate these phenomena.
6. identify the products formed from the hydrolysis of a given disaccharide.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• 1,4′ link
• disaccharide (see Section 25.1)
• invert sugar
Study Notes
Notice that most of the disaccharides discussed in this section contain one unit of D-glucose. You are not expected to remember the detailed structures of maltose, lactose and sucrose. Similarly, we do not expect you to remember the systematic names of these substances.
Previously, you learned that monosaccharides can form cyclic structures by the reaction of the carbonyl group with an OH group. These cyclic molecules can in turn react with another alcohol. Disaccharides (C12H22O11) are sugars composed of two monosaccharide units that are joined by a carbon–oxygen-carbon linkage known as a glycosidic linkage. This linkage is formed from the reaction of the anomeric carbon of one cyclic monosaccharide with the OH group of a second monosaccharide.
The disaccharides differ from one another in their monosaccharide constituents and in the specific type of glycosidic linkage connecting them. There are three common disaccharides: maltose, lactose, and sucrose. All three are white crystalline solids at room temperature and are soluble in water. We’ll consider each sugar in more detail.
Maltose
Maltose occurs to a limited extent in sprouting grain. It is formed most often by the partial hydrolysis of starch and glycogen. In the manufacture of beer, maltose is liberated by the action of malt (germinating barley) on starch; for this reason, it is often referred to as malt sugar. Maltose is about 30% as sweet as sucrose. The human body is unable to metabolize maltose or any other disaccharide directly from the diet because the molecules are too large to pass through the cell membranes of the intestinal wall. Therefore, an ingested disaccharide must first be broken down by hydrolysis into its two constituent monosaccharide units. In the body, such hydrolysis reactions are catalyzed by enzymes such as maltase. The same reactions can be carried out in the laboratory with dilute acid as a catalyst, although in that case the rate is much slower, and high temperatures are required. Whether it occurs in the body or a glass beaker, the hydrolysis of maltose produces two molecules of D-glucose.
$\mathrm{maltose \xrightarrow{H^+\: or\: maltase} \textrm{2 D-glucose}}$
Maltose is a reducing sugar. Thus, its two glucose molecules must be linked in such a way as to leave one anomeric carbon that can open to form an aldehyde group. The glucose units in maltose are joined in a head-to-tail fashion through an α-linkage from the first carbon atom of one glucose molecule to the fourth carbon atom of the second glucose molecule (that is, an α-1,4-glycosidic linkage; see Figure 1). The bond from the anomeric carbon of the first monosaccharide unit is directed downward, which is why this is known as an α-glycosidic linkage. The OH group on the anomeric carbon of the second glucose can be in either the α or the β position, as shown in Figure 1.
Figure 1 An Equilibrium Mixture of Maltose Isomers
Lactose
Lactose is known as milk sugar because it occurs in the milk of humans, cows, and other mammals. In fact, the natural synthesis of lactose occurs only in mammary tissue, whereas most other carbohydrates are plant products. Human milk contains about 7.5% lactose, and cow’s milk contains about 4.5%. This sugar is one of the lowest ranking in terms of sweetness, being about one-sixth as sweet as sucrose. Lactose is produced commercially from whey, a by-product in the manufacture of cheese. It is important as an infant food and in the production of penicillin.
Lactose is a reducing sugar composed of one molecule of D-galactose and one molecule of D-glucose joined by a β-1,4-glycosidic bond (the bond from the anomeric carbon of the first monosaccharide unit being directed upward). The two monosaccharides are obtained from lactose by acid hydrolysis or the catalytic action of the enzyme lactase:
Many adults and some children suffer from a deficiency of lactase. These individuals are said to be lactose intolerant because they cannot digest the lactose found in milk. A more serious problem is the genetic disease galactosemia, which results from the absence of an enzyme needed to convert galactose to glucose. Certain bacteria can metabolize lactose, forming lactic acid as one of the products. This reaction is responsible for the “souring” of milk.
Example 1
For this trisaccharide, indicate whether each glycosidic linkage is α or β.
Solution
The glycosidic linkage between sugars 1 and 2 is β because the bond is directed up from the anomeric carbon. The glycosidic linkage between sugars 2 and 3 is α because the bond is directed down from the anomeric carbon.
To Your Health: Lactose Intolerance and Galactosemia
Lactose makes up about 40% of an infant’s diet during the first year of life. Infants and small children have one form of the enzyme lactase in their small intestines and can digest the sugar easily; however, adults usually have a less active form of the enzyme, and about 70% of the world’s adult population has some deficiency in its production. As a result, many adults experience a reduction in the ability to hydrolyze lactose to galactose and glucose in their small intestine. For some people the inability to synthesize sufficient enzyme increases with age. Up to 20% of the US population suffers some degree of lactose intolerance.
In people with lactose intolerance, some of the unhydrolyzed lactose passes into the colon, where it tends to draw water from the interstitial fluid into the intestinal lumen by osmosis. At the same time, intestinal bacteria may act on the lactose to produce organic acids and gases. The buildup of water and bacterial decay products leads to abdominal distention, cramps, and diarrhea, which are symptoms of the condition.
The symptoms disappear if milk or other sources of lactose are excluded from the diet or consumed only sparingly. Alternatively, many food stores now carry special brands of milk that have been pretreated with lactase to hydrolyze the lactose. Cooking or fermenting milk causes at least partial hydrolysis of the lactose, so some people with lactose intolerance are still able to enjoy cheese, yogurt, or cooked foods containing milk. The most common treatment for lactose intolerance, however, is the use of lactase preparations (e.g., Lactaid), which are available in liquid and tablet form at drugstores and grocery stores. These are taken orally with dairy foods—or may be added to them directly—to assist in their digestion.
Galactosemia is a condition in which one of the enzymes needed to convert galactose to glucose is missing. Consequently, the blood galactose level is markedly elevated, and galactose is found in the urine. An infant with galactosemia experiences a lack of appetite, weight loss, diarrhea, and jaundice. The disease may result in impaired liver function, cataracts, mental retardation, and even death. If galactosemia is recognized in early infancy, its effects can be prevented by the exclusion of milk and all other sources of galactose from the diet. As a child with galactosemia grows older, he or she usually develops an alternate pathway for metabolizing galactose, so the need to restrict milk is not permanent. The incidence of galactosemia in the United States is 1 in every 65,000 newborn babies.
Sucrose
Sucrose, probably the largest-selling pure organic compound in the world, is known as beet sugar, cane sugar, table sugar, or simply sugar. Most of the sucrose sold commercially is obtained from sugar cane and sugar beets (whose juices are 14%–20% sucrose) by evaporation of the water and recrystallization. The dark brown liquid that remains after the recrystallization of sugar is sold as molasses.
The sucrose molecule is unique among the common disaccharides in having an α-1,β-2-glycosidic (head-to-head) linkage. Because this glycosidic linkage is formed by the OH group on the anomeric carbon of α-D-glucose and the OH group on the anomeric carbon of β-D-fructose, it ties up the anomeric carbons of both glucose and fructose.
This linkage gives sucrose certain properties that are quite different from those of maltose and lactose. As long as the sucrose molecule remains intact, neither monosaccharide “uncyclizes” to form an open-chain structure. Thus, sucrose is incapable of mutarotation and exists in only one form both in the solid state and in solution. In addition, sucrose does not undergo reactions that are typical of aldehydes and ketones. Therefore, sucrose is a nonreducing sugar.
The hydrolysis of sucrose in dilute acid or through the action of the enzyme sucrase (also known as invertase) gives an equimolar mixture of glucose and fructose. This 1:1 mixture is referred to as invert sugar because it rotates plane-polarized light in the opposite direction than sucrose. The hydrolysis reaction has several practical applications. Sucrose readily recrystallizes from a solution, but invert sugar has a much greater tendency to remain in solution. In the manufacture of jelly and candy and in the canning of fruit, the recrystallization of sugar is undesirable. Therefore, conditions leading to the hydrolysis of sucrose are employed in these processes. Moreover, because fructose is sweeter than sucrose, the hydrolysis adds to the sweetening effect. Bees carry out this reaction when they make honey.
The average American consumes more than 100 lb of sucrose every year. About two-thirds of this amount is ingested in soft drinks, presweetened cereals, and other highly processed foods. The widespread use of sucrose is a contributing factor to obesity and tooth decay. Carbohydrates such as sucrose, are converted to fat when the caloric intake exceeds the body’s requirements, and sucrose causes tooth decay by promoting the formation of plaque that sticks to teeth.
Summary
Maltose is composed of two molecules of glucose joined by an α-1,4-glycosidic linkage. It is a reducing sugar that is found in sprouting grain. Lactose is composed of a molecule of galactose joined to a molecule of glucose by a β-1,4-glycosidic linkage. It is a reducing sugar that is found in milk. Sucrose is composed of a molecule of glucose joined to a molecule of fructose by an α-1,β-2-glycosidic linkage. It is a nonreducing sugar that is found in sugar cane and sugar beets.
Concept Review Exercise
1. What monosaccharides are obtained by the hydrolysis of each disaccharide?
1. sucrose
2. maltose
3. lactose
Answer
1. D-glucose and D-fructose
2. two molecules of D-glucose
3. D-glucose and D-galactose
Exercises
1. Identify each sugar by its common chemical name.
1. milk sugar
2. table sugar
2. For each disaccharide, indicate whether the glycosidic linkage is α or β.
3. Identify each disaccharide in Exercise 2 as a reducing or nonreducing sugar. If it is a reducing sugar, draw its structure and circle the anomeric carbon. State if the OH group at the anomeric carbon is in the α or the β position
4. Melibiose is a disaccharide that occurs in some plant juices. Its structure is as follows:
1. What monosaccharide units are incorporated into melibiose?
2. What type of linkage (α or β) joins the two monosaccharide units of melibiose?
3. Melibiose has a free anomeric carbon and is thus a reducing sugar. Circle the anomeric carbon and indicate whether the OH group is α or β
Answers
1. lactose
2. sucrose
2. a.
2. b.
3. a. nonreducing
3. b. reducing
4. a. galactose and glucose
4. b.α-glycosidic linkage
4. c. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/24%3A_Carbohydrates%3A_Polyfunctional_Compounds_in_Nature/24.10%3A_Relative_Configurations__of_the_Aldoses%3A_An_Exercise__in_Structure__Determination.txt |
Objectives
After completing this section, you should be able to
1. identify the structural difference between cellulose and the cold‑water‑insoluble fraction of starch (amylose), and identify both of these substances as containing many glucose molecules joined by 1,4′‑glycoside links.
2. identify the cold‑water‑soluble fraction of starch (amylopectin) as having a more complex structure than amylose because of the existence of 1,6′‑glycoside links in addition to the 1,4′‑links.
3. compare and contrast the structures and uses of starch, glycogen and cellulose.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• amylopectin
• amylose
• polysaccharide
Learning Objectives
• To compare and contrast the structures and uses of starch, glycogen, and cellulose.
The polysaccharides are the most abundant carbohydrates in nature and serve a variety of functions, such as energy storage or as components of plant cell walls. Polysaccharides are very large polymers composed of tens to thousands of monosaccharides joined together by glycosidic linkages. The three most abundant polysaccharides are starch, glycogen, and cellulose. These three are referred to as homopolymers because each yields only one type of monosaccharide (glucose) after complete hydrolysis. Heteropolymers may contain sugar acids, amino sugars, or noncarbohydrate substances in addition to monosaccharides. Heteropolymers are common in nature (gums, pectins, and other substances) but will not be discussed further in this textbook. The polysaccharides are nonreducing carbohydrates, are not sweet tasting, and do not undergo mutarotation.
Starch
Starch is the most important source of carbohydrates in the human diet and accounts for more than 50% of our carbohydrate intake. It occurs in plants in the form of granules, and these are particularly abundant in seeds (especially the cereal grains) and tubers, where they serve as a storage form of carbohydrates. The breakdown of starch to glucose nourishes the plant during periods of reduced photosynthetic activity. We often think of potatoes as a “starchy” food, yet other plants contain a much greater percentage of starch (potatoes 15%, wheat 55%, corn 65%, and rice 75%). Commercial starch is a white powder.
Starch is a mixture of two polymers: amylose and amylopectin. Natural starches consist of about 10%–30% amylose and 70%–90% amylopectin. Amylose is a linear polysaccharide composed entirely of D-glucose units joined by the α-1,4-glycosidic linkages we saw in maltose (part (a) of Figure \(1\)). Experimental evidence indicates that amylose is not a straight chain of glucose units but instead is coiled like a spring, with six glucose monomers per turn (part (b) of Figure \(1\)). When coiled in this fashion, amylose has just enough room in its core to accommodate an iodine molecule. The characteristic blue-violet color that appears when starch is treated with iodine is due to the formation of the amylose-iodine complex. This color test is sensitive enough to detect even minute amounts of starch in solution.
Amylopectin is a branched-chain polysaccharide composed of glucose units linked primarily by α-1,4-glycosidic bonds but with occasional α-1,6-glycosidic bonds, which are responsible for the branching. A molecule of amylopectin may contain many thousands of glucose units with branch points occurring about every 25–30 units (Figure \(2\)). The helical structure of amylopectin is disrupted by the branching of the chain, so instead of the deep blue-violet color amylose gives with iodine, amylopectin produces a less intense reddish brown.
Dextrins are glucose polysaccharides of intermediate size. The shine and stiffness imparted to clothing by starch are due to the presence of dextrins formed when clothing is ironed. Because of their characteristic stickiness with wetting, dextrins are used as adhesives on stamps, envelopes, and labels; as binders to hold pills and tablets together; and as pastes. Dextrins are more easily digested than starch and are therefore used extensively in the commercial preparation of infant foods.
The complete hydrolysis of starch yields, in successive stages, glucose:
starch → dextrins → maltose → glucose
In the human body, several enzymes known collectively as amylases degrade starch sequentially into usable glucose units.
Glycogen
Glycogen is the energy reserve carbohydrate of animals. Practically all mammalian cells contain some stored carbohydrates in the form of glycogen, but it is especially abundant in the liver (4%–8% by weight of tissue) and in skeletal muscle cells (0.5%–1.0%). Like starch in plants, glycogen is found as granules in liver and muscle cells. When fasting, animals draw on these glycogen reserves during the first day without food to obtain the glucose needed to maintain metabolic balance.
Glycogen is structurally quite similar to amylopectin, although glycogen is more highly branched (8–12 glucose units between branches) and the branches are shorter. When treated with iodine, glycogen gives a reddish brown color. Glycogen can be broken down into its D-glucose subunits by acid hydrolysis or by the same enzymes that catalyze the breakdown of starch. In animals, the enzyme phosphorylase catalyzes the breakdown of glycogen to phosphate esters of glucose.
About 70% of the total glycogen in the body is stored in muscle cells. Although the percentage of glycogen (by weight) is higher in the liver, the much greater mass of skeletal muscle stores a greater total amount of glycogen.
Cellulose
Cellulose, a fibrous carbohydrate found in all plants, is the structural component of plant cell walls. Because the earth is covered with vegetation, cellulose is the most abundant of all carbohydrates, accounting for over 50% of all the carbon found in the vegetable kingdom. Cotton fibrils and filter paper are almost entirely cellulose (about 95%), wood is about 50% cellulose, and the dry weight of leaves is about 10%–20% cellulose. The largest use of cellulose is in the manufacture of paper and paper products. Although the use of noncellulose synthetic fibers is increasing, rayon (made from cellulose) and cotton still account for over 70% of textile production.
Like amylose, cellulose is a linear polymer of glucose. It differs, however, in that the glucose units are joined by β-1,4-glycosidic linkages, producing a more extended structure than amylose (part (a) of Figure \(3\)). This extreme linearity allows a great deal of hydrogen bonding between OH groups on adjacent chains, causing them to pack closely into fibers (part (b) of Figure \(3\)). As a result, cellulose exhibits little interaction with water or any other solvent. Cotton and wood, for example, are completely insoluble in water and have considerable mechanical strength. Because cellulose does not have a helical structure, it does not bind to iodine to form a colored product.
Cellulose yields D-glucose after complete acid hydrolysis, yet humans are unable to metabolize cellulose as a source of glucose. Our digestive juices lack enzymes that can hydrolyze the β-glycosidic linkages found in cellulose, so although we can eat potatoes, we cannot eat grass. However, certain microorganisms can digest cellulose because they make the enzyme cellulase, which catalyzes the hydrolysis of cellulose. The presence of these microorganisms in the digestive tracts of herbivorous animals (such as cows, horses, and sheep) allows these animals to degrade the cellulose from plant material into glucose for energy. Termites also contain cellulase-secreting microorganisms and thus can subsist on a wood diet. This example once again demonstrates the extreme stereospecificity of biochemical processes.
Career Focus: Certified Diabetes Educator
Certified diabetes educators come from a variety of health professions, such as nursing and dietetics, and specialize in the education and treatment of patients with diabetes. A diabetes educator will work with patients to manage their diabetes. This involves teaching the patient to monitor blood sugar levels, make good food choices, develop and maintain an exercise program, and take medication, if required.
Summary
Starch is a storage form of energy in plants. It contains two polymers composed of glucose units: amylose (linear) and amylopectin (branched). Glycogen is a storage form of energy in animals. It is a branched polymer composed of glucose units. It is more highly branched than amylopectin. Cellulose is a structural polymer of glucose units found in plants. It is a linear polymer with the glucose units linked through β-1,4-glycosidic bonds.
Contributors and Attributions
Objectives
You may omit Section 25.11.
Carbohydrates are covalently attached to many different biomolecules, including lipids, to form glycolipids, and proteins, to form glycoproteins. Glycoproteins and glycolipids are often found in biological membranes, to which they are anchored by through nonpolar interactions. A special kind of glycoprotein, a proteoglycan, actually has more carbohydrate mass than protein. What is the function of these carbohydrates? Two are apparent. First, glycosylation of proteins helps protect the protein from degradation by enzyme catalysts within the body. However, there main functions arises from the fact that covalently attached carbohydrates that "decorate" the surface of glycoproteins or glycolipids provide new binding site interactions that allow interactions with other biomolecules. Hence glycosylation allows for cell:cell, cell:protein, or protein:protein interactions. Unfortunately, bacteria and viruses often recognize glycosylated molecules on cell membranes, allowing for their import into the cell.
Here are some "cartoon" examples of carbohydrates covalently linked to the amino acid asparagine (Asn) on a glycoprotein.
Here are some examples of biomolecular interactions promoted by IMFs involving carbohydrates.
Influenza Virus binding to Cell Surface Glycoproteins with Neu5Ac - A protein on the surface of influenza virus, hemagluttinin, bind to sialic acid (Sia), which is covalently attached to many cell membrane glycoproteins on host cells. The sialic acid is usually connected through an alpha (2,3) or alpha (2,6) link to galactose on N-linked glycoproteins. The subtypes found in avian (and equine) influenza isolates bind preferentially to Sia (alpha 2,3) Gal which predominates in avian GI tract where viruses replicate. Human virus of H1, H2, and H3 subtype (cause of the 1918, 1957, and 1968 pandemics) recognize Sia (alpha 2,6) Gal, the major form in human respiratory tract. The swine influenza HA bind to Sia (alpha 2,6) Gal and some Sia (alpha 2,3)both of which found in swine.
Leukocyte: Cell Wall binding - During inflammation, circulating leukocytes (a type of white blood cell) tether and roll on the walls of blood vessels where they become active. E-, L- and P-selectin proteins are the primary proteins responsible for the tethering and rolling of these leukocytes. P-selectin binds, in part, to a tetrasaccharide, sialyl-Lewisx (SLEX) on the cell surface.. The interaction between P-selectin and the cell mediates the initial binding/rolling of the leukocyte on the vessel wall.
25: Heterocycles: Heteroatoms in Cyclic Organic Compounds
Thumbnail: Ball-and-stick model of the tetrahydrofuran molecule. (Public Domain; Ben Mills). | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/24%3A_Carbohydrates%3A_Polyfunctional_Compounds_in_Nature/24.11%3A_%09Complex__Sugars__in_Nature%3A_Disaccharides.txt |
Objectives
After completing this section, you should be able to
1. identify the structural features present in the 20 amino acids commonly found in proteins.
Note: You are not expected to remember the detailed structures of all these amino acids, but you should be prepared to draw the structures of the two simplest members, glycine and alanine.
2. draw the Fischer projection formula of a specified enantiomer of a given amino acid.
Note: To do so, you must remember that in the S enantiomer, the carboxyl group appears at the top of the projection formula and the amino group is on the left.
3. classify an amino acid as being acidic, basic or neutral, given its Kekulé, condensed or shorthand structure.
4. draw the zwitterion form of a given amino acid.
5. account for some of the typical properties of amino acids (e.g., high melting points, solubility in water) in terms of zwitterion formation.
6. write appropriate equations to illustrate the amphoteric nature of amino acids.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• α‑amino acids
• amphoteric
• essential amino acids
• zwitterion
Study Notes
This is a good point at which to review some of the principles of stereochemistry presented in Chapter 5. Be sure to make full use of molecular models when any stereochemical issues arise.
You should recognize that a three‑letter shorthand code is often used to represent individual amino acids. You need not memorize this code.
The distinction between essential and nonessential amino acids is not as clear‑cut as one might suppose. For example, arginine is often regarded as being nonessential.
Introduction to Amino Acids
Amino acids form polymers through a nucleophilic attack by the amino group of an amino acid at the electrophilic carbonyl carbon of the carboxyl group of another amino acid. The carboxyl group of the amino acid must first be activated to provide a better leaving group than OH-. (We will discuss this activation by ATP later in the course.) The resulting link between the amino acids is an amide link which biochemists call a peptide bond. In this reaction, water is released. In a reverse reaction, the peptide bond can be cleaved by water (hydrolysis).
When two amino acids link together to form an amide link, the resulting structure is called a dipeptide. Likewise, we can have tripeptides, tetrapeptides, and other polypeptides. At some point, when the structure is long enough, it is called a protein. There are many different ways to represent the structure of a polypeptide or protein, each showing differing amounts of information.
Figure: Different Representations of a Polypeptide (Heptapeptide)
Figure: Amino Acids React to Form Proteins
(Note: above picture represents the amino acid in an unlikely protonation state with the weak acid protonated and the weak base deprotonated for simplicity in showing removal of water on peptide bond formation and the hydrolysis reaction.) Proteins are polymers of twenty naturally occurring amino acids. In contrast, nucleic acids are polymers of just 4 different monomeric nucleotides. Both the sequence of a protein and its total length differentiate one protein from another. Just for an octapeptide, there are over 25 billion different possible arrangements of amino acids. Compare this to just 65536 different oligonucleotides of 8 monomeric units (8mer). Hence the diversity of possible proteins is enormous.
Stereochemistry
The amino acids are all chiral, with the exception of glycine, whose side chain is H. As with lipids, biochemists use the L and D nomenclature. All naturally occuring proteins from all living organisms consist of L amino acids. The absolute stereochemistry is related to L-glyceraldehyde, as was the case for triacylglycerides and phospholipids. Most naturally occurring chiral amino acids are S, with the exception of cysteine. As the diagram below shows, the absolute configuration of the amino acids can be shown with the H pointed to the rear, the COOH groups pointing out to the left, the R group to the right, and the NH3 group upwards. You can remember this with the anagram CORN.
Figure: Stereochemistry of Amino Acids.
Why do biochemists still use D and L for sugars and amino acids? This explanation (taken from the link below) seems reasonable.
"In addition, however, chemists often need to define a configuration unambiguously in the absence of any reference compound, and for this purpose the alternative (R,S) system is ideal, as it uses priority rules to specify configurations. These rules sometimes lead to absurd results when they are applied to biochemical molecules. For example, as we have seen, all of the common amino acids are L, because they all have exactly the same structure, including the position of the R group if we just write the R group as R. However, they do not all have the same configuration in the (R,S) system: L-cysteine is also (R)-cysteine, but all the other L-amino acids are (S), but this just reflects the human decision to give a sulphur atom higher priority than a carbon atom, and does not reflect a real difference in configuration. Worse problems can sometimes arise in substitution reactions: sometimes inversion of configuration can result in no change in the (R) or (S) prefix; and sometimes retention of configuration can result in a change of prefix.
It follows that it is not just conservatism or failure to understand the (R,S) system that causes biochemists to continue with D and L: it is just that the DL system fulfils their needs much better. As mentioned, chemists also use D and L when they are appropriate to their needs. The explanation given above of why the (R,S) system is little used in biochemistry is thus almost the exact opposite of reality. This system is actually the only practical way of unambiguously representing the stereochemistry of complicated molecules with several asymmetric centres, but it is inconvenient with regular series of molecules like amino acids and simple sugars. "
Natural α-Amino Acids
Hydrolysis of proteins by boiling aqueous acid or base yields an assortment of small molecules identified as α-aminocarboxylic acids. More than twenty such components have been isolated, and the most common of these are listed in the following table. Those amino acids having green colored names are essential diet components, since they are not synthesized by human metabolic processes. The best food source of these nutrients is protein, but it is important to recognize that not all proteins have equal nutritional value. For example, peanuts have a higher weight content of protein than fish or eggs, but the proportion of essential amino acids in peanut protein is only a third of that from the two other sources. For reasons that will become evident when discussing the structures of proteins and peptides, each amino acid is assigned a one or three letter abbreviation.
Natural α-Amino Acids
Some common features of these amino acids should be noted. With the exception of proline, they are all 1º-amines; and with the exception of glycine, they are all chiral. The configurations of the chiral amino acids are the same when written as a Fischer projection formula, as in the drawing on the right, and this was defined as the L-configuration by Fischer. The R-substituent in this structure is the remaining structural component that varies from one amino acid to another, and in proline R is a three-carbon chain that joins the nitrogen to the alpha-carbon in a five-membered ring. Applying the Cahn-Ingold-Prelog notation, all these natural chiral amino acids, with the exception of cysteine, have an S-configuration. For the first seven compounds in the left column the R-substituent is a hydrocarbon. The last three entries in the left column have hydroxyl functional groups, and the first two amino acids in the right column incorporate thiol and sulfide groups respectively. Lysine and arginine have basic amine functions in their side-chains; histidine and tryptophan have less basic nitrogen heterocyclic rings as substituents. Finally, carboxylic acid side-chains are substituents on aspartic and glutamic acid, and the last two compounds in the right column are their corresponding amides.
The formulas for the amino acids written above are simple covalent bond representations based upon previous understanding of mono-functional analogs. The formulas are in fact incorrect. This is evident from a comparison of the physical properties listed in the following table. All four compounds in the table are roughly the same size, and all have moderate to excellent water solubility. The first two are simple carboxylic acids, and the third is an amino alcohol. All three compounds are soluble in organic solvents (e.g. ether) and have relatively low melting points. The carboxylic acids have pKa's near 4.5, and the conjugate acid of the amine has a pKa of 10. The simple amino acid alanine is the last entry. By contrast, it is very high melting (with decomposition), insoluble in organic solvents, and a million times weaker as an acid than ordinary carboxylic acids.
Physical Properties of Selected Acids and Amines
Compound
Formula
Mol.Wt.
Solubility in Water
Solubility in Ether
Melting Point
pKa
isobutyric acid (CH3)2CHCO2H 88 20g/100mL complete -47 ºC 5.0
lactic acid CH3CH(OH)CO2H 90 complete complete 53 ºC 3.9
3-amino-2-butanol CH3CH(NH2)CH(OH)CH3 89 complete complete 9 ºC 10.0
alanine CH3CH(NH2)CO2H 89 18g/100mL insoluble ca. 300 ºC 9.8
Zwitterion
These differences above all point to internal salt formation by a proton transfer from the acidic carboxyl function to the basic amino group. The resulting ammonium carboxylate structure, commonly referred to as a zwitterion, is also supported by the spectroscopic characteristics of alanine.
CH3CH(NH2)CO2H CH3CH(NH3)(+)CO2(–)
As expected from its ionic character, the alanine zwitterion is high melting, insoluble in nonpolar solvents and has the acid strength of a 1º-ammonium ion. Examples of a few specific amino acids may also be viewed in their favored neutral zwitterionic form. Note that in lysine the amine function farthest from the carboxyl group is more basic than the alpha-amine. Consequently, the positively charged ammonium moiety formed at the chain terminus is attracted to the negative carboxylate, resulting in a coiled conformation.
The structure of an amino acid allows it to act as both an acid and a base. An amino acid has this ability because at a certain pH value (different for each amino acid) nearly all the amino acid molecules exist as zwitterions. If acid is added to a solution containing the zwitterion, the carboxylate group captures a hydrogen (H+) ion, and the amino acid becomes positively charged. If base is added, ion removal of the H+ ion from the amino group of the zwitterion produces a negatively charged amino acid. In both circumstances, the amino acid acts to maintain the pH of the system—that is, to remove the added acid (H+) or base (OH) from solution.
Example 26.1
1. Draw the structure for the anion formed when glycine (at neutral pH) reacts with a base.
2. Draw the structure for the cation formed when glycine (at neutral pH) reacts with an acid.
Solution
1. The base removes H+ from the protonated amine group.
2. The acid adds H+ to the carboxylate group.
Other Natural Amino Acids
The twenty alpha-amino acids listed above are the primary components of proteins, their incorporation being governed by the genetic code. Many other naturally occurring amino acids exist, and the structures of a few of these are displayed below. Some, such as hydroxylysine and hydroxyproline, are simply functionalized derivatives of a previously described compound. These two amino acids are found only in collagen, a common structural protein. Homoserine and homocysteine are higher homologs of their namesakes. The amino group in beta-alanine has moved to the end of the three-carbon chain. It is a component of pantothenic acid, HOCH2C(CH3)2CH(OH)CONHCH2CH2CO2H, a member of the vitamin B complex and an essential nutrient. Acetyl coenzyme A is a pyrophosphorylated derivative of a pantothenic acid amide. The gamma-amino homolog GABA is a neurotransmitter inhibitor and antihypertensive agent.
Many unusual amino acids, including D-enantiomers of some common acids, are produced by microorganisms. These include ornithine, which is a component of the antibiotic bacitracin A, and statin, found as part of a pentapeptide that inhibits the action of the digestive enzyme pepsin. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/26%3A_Amino_Acids_Peptides_Proteins_and_Nucleic_Acids%3A_Nitrogen-Containing_Polymers_in_Nature/26.01%3A_Structure_and__Properties_of_Amino__Acids.txt |
Objectives
After completing this section, you should be able to
1. outline, by means of equations, how a racemic mixture of given amino acid can be prepared from a carboxylic acid using reactions you studied earlier in the course.
1. outline, by means of equations, the preparation of a given amino acid by the amidomalonate synthesis.
2. identify the amino acid formed from using a given alkyl halide in an amidomalonate synthesis.
3. identify the alkyl halide needed to produce a given amino acid by the amidomalonate synthesis.
2. describe, by means of equations, how an α‑keto acid can be transformed to an amino acid by reductive amination.
1. describe a general method for resolving a racemic mixture of a given amino acid.
2. provide a brief example of how a biological method may be employed to resolve a racemic mixture of a given amino acid.
3. show the enantioselective preparation of an amino acid from the corresponding Z enamido acid.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• amidomalonate synthesis
• enantioselective synthesis
• racemic mixture
Study Notes
Do not be alarmed by the number of methods to synthesize amino acids described in this section.You have seen many of these reactions in previous sections and should already be familiar with the approaches discussed here.
To fulfill the requirements of Objective 1, review the Hell‑Volhard‑Zelinskii reaction (Section 22.4) and the Gabriel phthalimide synthesis (Section 24.6).
The amidomalonate synthesis is a simple variation of the malonic ester synthesis (Section 22.7). A base abstracts a proton from the alpha carbon, which is then alkylated with an alkyl halide. Then both the hydrolysis of the esters and the amide protecting group under aqueous acidic conditions generates the α‑amino acid.
Another method of getting to the α‑amino acid is by reductive amination of the α‑keto acid which you have also previously encountered (Section 24.6).
Synthesis of α-Amino Acids
1) Amination of alpha-bromocarboxylic acids, illustrated by the following equation, provides a straightforward method for preparing alpha-aminocarboxylic acids. The bromoacids, in turn, are conveniently prepared from carboxylic acids by reaction with Br2 + PCl3. Although this direct approach gave mediocre results when used to prepare simple amines from alkyl halides, it is more effective for making amino acids, thanks to the reduced nucleophilicity of the nitrogen atom in the product. Nevertheless, more complex procedures that give good yields of pure compounds are often chosen for amino acid synthesis.
2) By modifying the nitrogen as a phthalimide salt, the propensity of amines to undergo multiple substitutions is removed, and a single clean substitution reaction of 1º- and many 2º-alkylhalides takes place. This procedure, known as the Gabriel synthesis, can be used to advantage in aminating bromomalonic esters, as shown in the upper equation of the following scheme. Since the phthalimide substituted malonic ester has an acidic hydrogen (colored orange), activated by the two ester groups, this intermediate may be converted to an ambident anion and alkylated. Finally, base catalyzed hydrolysis of the phthalimide moiety and the esters, followed by acidification and thermal decarboxylation, produces an amino acid and phthalic acid (not shown).
3) An elegant procedure, known as the Strecker synthesis, assembles an alpha-amino acid from ammonia (the amine precursor), cyanide (the carboxyl precursor), and an aldehyde. This reaction (shown below) is essentially an imino analog of cyanohydrin formation. The alpha-amino nitrile formed in this way can then be hydrolyzed to an amino acid by either acid or base catalysis.
4) Resolution The three synthetic procedures described above, and many others that can be conceived, give racemic amino acid products. If pure L or D enantiomers are desired, it is necessary to resolve these racemic mixtures. A common method of resolving racemates is by diastereomeric salt formation with a pure chiral acid or base. This is illustrated for a generic amino acid in the following diagram. Be careful to distinguish charge symbols, shown in colored circles, from optical rotation signs, shown in parenthesis.
In the initial display, the carboxylic acid function contributes to diastereomeric salt formation. The racemic amino acid is first converted to a benzamide derivative to remove the basic character of the amino group. Next, an ammonium salt is formed by combining the carboxylic acid with an optically pure amine, such as brucine (a relative of strychnine). The structure of this amine is not shown, because it is not a critical factor in the logical progression of steps. Since the amino acid moiety is racemic and the base is a single enantiomer (levorotatory in this example), an equimolar mixture of diastereomeric salts is formed (drawn in the green shaded box). Diastereomers may be separated by crystallization, chromatography or other physical manipulation, and in this way one of the isomers may be isolated for further treatment, in this illustration it is the (+):(-) diastereomer. Finally the salt is broken by acid treatment, giving the resolved (+)-amino acid derivative together with the recovered resolving agent (the optically active amine). Of course, the same procedure could be used to obtain the (-)-enantiomer of the amino acid.
Since amino acids are amphoteric, resolution could also be achieved by using the basic character of the amine function. For this approach we would need an enantiomerically pure chiral acid such as tartaric acid to use as the resolving agent. This alternative resolution strategy will be illustrated. Note that the carboxylic acid function is first esterified, so that it will not compete with the resolving acid.
Resolution of aminoacid derivatives may also be achieved by enzymatic discrimination in the hydrolysis of amides. For example, an aminoacylase enzyme from pig kidneys cleaves an amide derivative of a natural L-amino acid much faster than it does the D-enantiomer. If the racemic mixture of amides shown in the green shaded box above is treated with this enzyme, the L-enantiomer (whatever its rotation) will be rapidly converted to its free zwitterionic form, whereas the D-enantiomer will remain largely unchanged. Here, the diastereomeric species are transition states rather than isolable intermediates. This separation of enantiomers, based on very different rates of reaction, is called kinetic resolution.
Enantioselective Synthesis
Till now all of the synthetic routes to α-amino acids we have discussed yield a racemic mixture. Once produced one could resolve the mixture to obtain pure L or D enantiomers. However, enantioselective synthetic methods to produce pure compounds directly are being developed. For instance, several catalysts are now available for reduction of C=C to expose enantiopure amino acids. A good example is the industrial synthesis of L-DOPA, a drug used in the treatment of Parkinson’s disease. W.S. Knowles shared the 2001 Nobel Price with R. Noyori and K.B. Sharpless for their contributions in the area of asymmetric catalytic reductions. Knowles developed several chiral phosphine–metal catalysts for asymmetric reductions. The rhodium(I) catalyst shown, which is complexed by large organic ligands, facilitates production of almost pure L-DOPA. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/26%3A_Amino_Acids_Peptides_Proteins_and_Nucleic_Acids%3A_Nitrogen-Containing_Polymers_in_Nature/26.02%3A_Synthesis_of_Amino__Acids%3A_A_Combination_of_Amine.txt |
Objectives
After completing this section, you should be able to
1. outline, by means of equations, how a racemic mixture of given amino acid can be prepared from a carboxylic acid using reactions you studied earlier in the course.
1. outline, by means of equations, the preparation of a given amino acid by the amidomalonate synthesis.
2. identify the amino acid formed from using a given alkyl halide in an amidomalonate synthesis.
3. identify the alkyl halide needed to produce a given amino acid by the amidomalonate synthesis.
2. describe, by means of equations, how an α‑keto acid can be transformed to an amino acid by reductive amination.
1. describe a general method for resolving a racemic mixture of a given amino acid.
2. provide a brief example of how a biological method may be employed to resolve a racemic mixture of a given amino acid.
3. show the enantioselective preparation of an amino acid from the corresponding Z enamido acid.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• amidomalonate synthesis
• enantioselective synthesis
• racemic mixture
Study Notes
Do not be alarmed by the number of methods to synthesize amino acids described in this section.You have seen many of these reactions in previous sections and should already be familiar with the approaches discussed here.
To fulfill the requirements of Objective 1, review the Hell‑Volhard‑Zelinskii reaction (Section 22.4) and the Gabriel phthalimide synthesis (Section 24.6).
The amidomalonate synthesis is a simple variation of the malonic ester synthesis (Section 22.7). A base abstracts a proton from the alpha carbon, which is then alkylated with an alkyl halide. Then both the hydrolysis of the esters and the amide protecting group under aqueous acidic conditions generates the α‑amino acid.
Another method of getting to the α‑amino acid is by reductive amination of the α‑keto acid which you have also previously encountered (Section 24.6).
Synthesis of α-Amino Acids
1) Amination of alpha-bromocarboxylic acids, illustrated by the following equation, provides a straightforward method for preparing alpha-aminocarboxylic acids. The bromoacids, in turn, are conveniently prepared from carboxylic acids by reaction with Br2 + PCl3. Although this direct approach gave mediocre results when used to prepare simple amines from alkyl halides, it is more effective for making amino acids, thanks to the reduced nucleophilicity of the nitrogen atom in the product. Nevertheless, more complex procedures that give good yields of pure compounds are often chosen for amino acid synthesis.
2) By modifying the nitrogen as a phthalimide salt, the propensity of amines to undergo multiple substitutions is removed, and a single clean substitution reaction of 1º- and many 2º-alkylhalides takes place. This procedure, known as the Gabriel synthesis, can be used to advantage in aminating bromomalonic esters, as shown in the upper equation of the following scheme. Since the phthalimide substituted malonic ester has an acidic hydrogen (colored orange), activated by the two ester groups, this intermediate may be converted to an ambident anion and alkylated. Finally, base catalyzed hydrolysis of the phthalimide moiety and the esters, followed by acidification and thermal decarboxylation, produces an amino acid and phthalic acid (not shown).
3) An elegant procedure, known as the Strecker synthesis, assembles an alpha-amino acid from ammonia (the amine precursor), cyanide (the carboxyl precursor), and an aldehyde. This reaction (shown below) is essentially an imino analog of cyanohydrin formation. The alpha-amino nitrile formed in this way can then be hydrolyzed to an amino acid by either acid or base catalysis.
4) Resolution The three synthetic procedures described above, and many others that can be conceived, give racemic amino acid products. If pure L or D enantiomers are desired, it is necessary to resolve these racemic mixtures. A common method of resolving racemates is by diastereomeric salt formation with a pure chiral acid or base. This is illustrated for a generic amino acid in the following diagram. Be careful to distinguish charge symbols, shown in colored circles, from optical rotation signs, shown in parenthesis.
In the initial display, the carboxylic acid function contributes to diastereomeric salt formation. The racemic amino acid is first converted to a benzamide derivative to remove the basic character of the amino group. Next, an ammonium salt is formed by combining the carboxylic acid with an optically pure amine, such as brucine (a relative of strychnine). The structure of this amine is not shown, because it is not a critical factor in the logical progression of steps. Since the amino acid moiety is racemic and the base is a single enantiomer (levorotatory in this example), an equimolar mixture of diastereomeric salts is formed (drawn in the green shaded box). Diastereomers may be separated by crystallization, chromatography or other physical manipulation, and in this way one of the isomers may be isolated for further treatment, in this illustration it is the (+):(-) diastereomer. Finally the salt is broken by acid treatment, giving the resolved (+)-amino acid derivative together with the recovered resolving agent (the optically active amine). Of course, the same procedure could be used to obtain the (-)-enantiomer of the amino acid.
Since amino acids are amphoteric, resolution could also be achieved by using the basic character of the amine function. For this approach we would need an enantiomerically pure chiral acid such as tartaric acid to use as the resolving agent. This alternative resolution strategy will be illustrated. Note that the carboxylic acid function is first esterified, so that it will not compete with the resolving acid.
Resolution of aminoacid derivatives may also be achieved by enzymatic discrimination in the hydrolysis of amides. For example, an aminoacylase enzyme from pig kidneys cleaves an amide derivative of a natural L-amino acid much faster than it does the D-enantiomer. If the racemic mixture of amides shown in the green shaded box above is treated with this enzyme, the L-enantiomer (whatever its rotation) will be rapidly converted to its free zwitterionic form, whereas the D-enantiomer will remain largely unchanged. Here, the diastereomeric species are transition states rather than isolable intermediates. This separation of enantiomers, based on very different rates of reaction, is called kinetic resolution.
Enantioselective Synthesis
Till now all of the synthetic routes to α-amino acids we have discussed yield a racemic mixture. Once produced one could resolve the mixture to obtain pure L or D enantiomers. However, enantioselective synthetic methods to produce pure compounds directly are being developed. For instance, several catalysts are now available for reduction of C=C to expose enantiopure amino acids. A good example is the industrial synthesis of L-DOPA, a drug used in the treatment of Parkinson’s disease. W.S. Knowles shared the 2001 Nobel Price with R. Noyori and K.B. Sharpless for their contributions in the area of asymmetric catalytic reductions. Knowles developed several chiral phosphine–metal catalysts for asymmetric reductions. The rhodium(I) catalyst shown, which is complexed by large organic ligands, facilitates production of almost pure L-DOPA. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/26%3A_Amino_Acids_Peptides_Proteins_and_Nucleic_Acids%3A_Nitrogen-Containing_Polymers_in_Nature/26.03%3A_Synthesis_of_Enantiomerically_Pure__Amino__Acids.txt |
Objectives
After completing this section, you should be able to
1. show, by means of a diagram, how two different amino acid residues can be combined to give two different dipeptides.
2. draw the structure of a relatively simple peptide, given its full or abbreviated name and the structures of the appropriate amino acids.
3. draw, or name, the six possible isomeric tripeptides that can be formed by combining three different amino acid residues (amino acid units) of given structure.
1. account for the fact that there is restricted rotation about the C$\ce{-}$N bonds in peptides.
2. illustrate the formation of a disulfide linkage between two cysteine residues, and show how such bonds can link together two separate peptide chains or can provide a bridge between two cysteine residues present in a single peptide molecule.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• C‑terminal amino acid
• N‑terminal amino acid
• peptides
• residues
Study Notes
If necessary, review the discussion of the delocalization of the nitrogen lone‑pair electrons in amides that was presented in Section 24.3. Similarly, you may wish to refer back to Section 18.8 to review the interconversion of thiols and disulfides.
Peptide Bond Formation or Amide Synthesis
The formation of peptides is nothing more than the application of the amide synthesis reaction. By convention, the amide bond in the peptides should be made in the order that the amino acids are written. The amine end (N terminal) of an amino acid is always on the left, while the acid end (C terminal) is on the right. The reaction of glycine with alanine to form the dipeptide glyclalanine is written as shown in the graphic on the left. Oxygen (red) from the acid and hydrogens (red) on the amine form a water molecule. The carboxyl oxygen (green) and the amine nitrogen (green) join to form the amide bond.
If the order of listing the amino acids is reversed, a different dipeptide is formed such as alaninylglycine.
Exercise $1$
Write the reactions for:
1. ala + gly ---> Answer graphic
2. phe + ser ----> Answer graphic
Resonance contributors for the peptide bonds
A consideration of resonance contributors is crucial to any discussion of the amide functional group. One of the most important examples of amide groups in nature is the ‘peptide bond’ that links amino acids to form polypeptides and proteins.
Critical to the structure of proteins is the fact that, although it is conventionally drawn as a single bond, the C-N bond in a peptide linkage has a significant barrier to rotation, almost as if it were a double bond.
This, along with the observation that the bonding around the peptide nitrogen has trigonal planar geometry, strongly suggests that the nitrogen is sp2-hybridized. An important resonance contributor has a C=N double bond and a C-O single bond, with a separation of charge between the oxygen and the nitrogen.
Although B is a minor contributor due to the separation of charges, it is still very relevant in terms of peptide and protein structure – our proteins would simply not fold up properly if there was free rotation about the peptide C-N bond.
Backbone Peptide or Protein Structure
The structure of a peptide can be written fairly easily without showing the complete amide synthesis reaction by learning the structure of the "backbone" for peptides and proteins.
The peptide backbone consists of repeating units of "N-H 2, CH, C double bond O; N-H 2, CH, C double bond O; etc. See the graphic on the left .
After the backbone is written, go back and write the specific structure for the side chains as represented by the "R" as gly-ala-leu for this example. The amine end (N terminal) of an amino acid is always on the left (gly), while the acid end (C terminal) is on the right (leu).
Exercise $2$
Write the tripeptide structure for val-ser-cys. First write the "backbone" and then add the specific side chains.
Solution
Answer graphic
QUES. Write the structure for the tripeptide:
2 a ) glu-cys-gly ---> Answer graphic
2 b) phe-tyr-asn ---> Answer graphic
Disulfide Bridges and Oxidation-Reduction
The amino acid cysteine undergoes oxidation and reduction reactions involving the -SH (sulfhydryl group). The oxidation of two sulfhydryl groups results in the formation of a disulfide bond by the removal of two hydrogens. The oxidation of two cysteine amino acids is shown in the graphic. An unspecified oxidizing agent (O) provides an oxygen which reacts with the hydrogen (red) on the -SH group to form water. The sulfurs (yellow) join to make the disulfide bridge. This is an important bond to recognize in protein tertiary structure. The reduction of a disulfide bond is the opposite reaction which again leads to two separate cysteine molecules. Remember that reduction is the addition of hydrogen.
Cysteine residues in the the peptide chain can form a loop buy forming the disulfide bond (—S—S—), while cysteine residues in different peptide chains can actually link what were otherwise separate chains. Insulin was the first protein whose amino acid sequence was determined. This pioneering work, completed in 1953 after some 10 years of effort, earned a Nobel Prize for British biochemist Frederick Sanger (born 1918). He found the primary structure to comprise of two chains linked by two cysteine disulfide bridges. Also note the first peptide chain possesses an internal loop.
Insulin
Contributors and Attributions
Charles Ophardt (Professor Emeritus, Elmhurst College); Virtual Chembook
Objectives
After completing this section, you should be able to
1. discuss, with reference to a suitable example (either given or of your own choice), the structure of proteins, paying particular attention to distinguishing between the primary, secondary, tertiary and quaternary structure.
2. describe the α‑helical secondary structure displayed by many proteins.
3. describe the β‑pleated‑sheet structure displayed by many proteins.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• α helix
• β pleated sheet
• primary structure
• quaternary structure
• secondary structure
• tertiary structure
Study Notes
Note that in a diagram of the α‑helical structure of a protein, the C‑terminal of the protein is at the bottom of the diagram and the N‑terminal is at the top. In an α helix, such as the one shown in Figure 26.9.1, the bulky R groups are all found on the outside of the helix, where they have the most room.
The four levels of protein structure
Protein structure can be discussed at four distinct levels. A protein’s primary structure is two-dimensional - simply the sequence of amino acids in the peptide chain. Below is a Lewis structure of a short segment of a protein with the sequence CHEM (cysteine - histidine - glutamate - methionine)
Secondary structure is three-dimensional, but is a local phenomenon, confined to a relatively short stretch of amino acids. For the most part, there are three important elements of secondary structure: helices, beta-sheets, and loops. In a helix, the main chain of the protein adopts the shape of a clockwise spiral staircase, and the side chains point out laterally.
In a beta-sheet (or beta-strand) structure, two sections of protein chain are aligned side-by-side in an extended conformation. The figure below shows two different views of the same beta-sheet: in the left-side view, the two regions of protein chain are differentiated by color.
Loops are relatively disordered segments of protein chain, but often assume a very ordered structure when in contact with a second protein or a smaller organic compound.
Both helix and the beta-sheet structures are held together by very specific hydrogen-bonding interactions between the amide nitrogen on one amino acid and the carbonyl oxygen on another. The hydrogen bonding pattern in a section of a beta-strand is shown below.
Secondary structure refers to the shape of a folding protein due exclusively to hydrogen bonding between its backbone amide and carbonyl groups. Secondary structure does not include bonding between the R-groups of amino acids, hydrophobic interactions, or other interactions associated with tertiary structure. The two most commonly encountered secondary structures of a polypeptide chain are α-helices and beta-pleated sheets. These structures are the first major steps in the folding of a polypeptide chain, and they establish important topological motifs that dictate subsequent tertiary structure and the ultimate function of the protein.
α-Helices
Figure 26.9.1 Ball-and-stick model of the α helix. Hydrogen bonds are shown as dotted bonds. Note that R groups extend almost perpendicular from the axis.
An α-helix is a right-handed coil of amino-acid residues on a polypeptide chain, typically ranging between 4 and 40 residues. This coil is held together by hydrogen bonds between the oxygen of C=O on top coil and the hydrogen of N-H on the bottom coil. Such a hydrogen bond is formed exactly every 4 amino acid residues, and every complete turn of the helix is only 3.6 amino acid residues. This regular pattern gives the α-helix very definite features with regards to the thickness of the coil and the length of each complete turn along the helix axis.
The structural integrity of an α-helix is in part dependent on correct steric configuration. Amino acids whose R-groups are too large (tryptophan, tyrosine) or too small (glycine) destabilize α-helices. Proline also destabilizes α-helices because of its irregular geometry; its R-group bonds back to the nitrogen of the amide group, which causes steric hindrance. In addition, the lack of a hydrogen on Proline's nitrogen prevents it from participating in hydrogen bonding.
Another factor affecting α-helix stability is the total dipole moment of the entire helix due to individual dipoles of the C=O groups involved in hydrogen bonding. Stable α-helices typically end with a charged amino acid to neutralize the dipole moment.
BETA-PLEATED SHEETS
This structure occurs when two (or more, e.g. ψ-loop) segments of a polypeptide chain overlap one another and form a row of hydrogen bonds with each other. This can happen in a parallel arrangement:
Or in anti-parallel arrangement:
Parallel and anti-parallel arrangement is the direct consequence of the directionality of the polypeptide chain. In anti-parallel arrangement, the C-terminus end of one segment is on the same side as the N-terminus end of the other segment. In parallel arrangement, the C-terminus end and the N-terminus end are on the same sides for both segments. The "pleat" occurs because of the alternating planes of the peptide bonds between amino acids; the aligned amino and carbonyl group of each opposite segment alternate their orientation from facing towards each other to facing opposite directions.
The parallel arrangement is less stable because the geometry of the individual amino acid molecules forces the hydrogen bonds to occur at an angle, making them longer and thus weaker. Contrarily, in the anti-parallel arrangement the hydrogen bonds are aligned directly opposite each other, making for stronger and more stable bonds.
Commonly, an anti-parallel beta-pleated sheet forms when a polypeptide chain sharply reverses direction. This can occur in the presence of two consecutive proline residues, which create an angled kink in the polypeptide chain and bend it back upon itself. This is not necessary for distant segments of a polypeptide chain to form beta-pleated sheets, but for proximal segments it is a definite requirement. For short distances, the two segments of a beta-pleated sheet are separated by 4+2n amino acid residues, with 4 being the minimum number of residues.
α-PLEATED SHEETS
A similar structure to the beta-pleated sheet is the α-pleated sheet. This structure is energetically less favorable than the beta-pleated sheet, and is fairly uncommon in proteins. An α-pleated sheet is characterized by the alignment of its carbonyl and amino groups; the carbonyl groups are all aligned in one direction, while all the N-H groups are aligned in the opposite direction. The polarization of the amino and carbonyl groups results in a net dipole moment on the α-pleated sheet. The carbonyl side acquires a net negative charge, and the amino side acquires a net positive charge.
A protein’s tertiary structure is the shape in which the entire protein chain folds together in three-dimensional space, and it is this level of structure that provides protein scientists with the most information about a protein’s specific function.
While a protein's secondary and tertiary structure is defined by how the protein chain folds together, quaternary structure is defined by how two or more folded protein chains come together to form a 'superstructure'. Many proteins consist of only one protein chain, or subunit, and thus have no quaternary structure. Many other proteins consist of two identical subunits (these are called homodimers) or two non-identical subunits (these are called heterodimers).
Quaternary structures can be quite elaborate: below we see a protein whose quaternary structure is defined by ten identical subunits arranged in two five-membered rings, forming what can be visualized as a 'double donut' shape (this is fructose 1,6-bisphosphate aldolase):
The molecular forces that hold proteins together
The question of exactly how a protein ‘finds’ its specific folded structure out of the vast number of possible folding patterns is still an active area of research. What is known, however, is that the forces that cause a protein to fold properly and to remain folded are the same basic noncovalent forces that we talked about in chapter 2: ion-ion, ion-dipole, dipole-dipole, hydrogen bonding, and hydrophobic (van der Waals) interactions. One interesting type of hydrophobic interaction is called ‘aromatic stacking’, and occurs when two or more planar aromatic rings on the side chains of phenylalanine, tryptophan, or tyrosine stack together like plates, thus maximizing surface area contact.
Hydrogen bonding networks are extensive within proteins, with both side chain and main chain atoms participating. Ionic interactions often play a role in protein structure, especially on the protein surface, as negatively charged residues such as aspartate interact with positively-charged groups on lysine or arginine.
One of the most important ideas to understand regarding tertiary structure is that a protein, when properly folded, is polar on the surface and nonpolar in the interior. It is the protein's surface that is in contact with water, and therefore the surface must be hydrophilic in order for the whole structure to be soluble. If you examine a three dimensional protein structure you will see many charged side chains (e.g. lysine, arginine, aspartate, glutamate) and hydrogen-bonding side chains (e.g. serine, threonine, glutamine, asparagine) exposed on the surface, in direct contact with water. Inside the protein, out of contact with the surrounding water, there tend to be many more hydrophobic residues such as alanine, valine, phenylalanine, etc. If a protein chain is caused to come unfolded (through exposure to heat, for example, or extremes of pH), it will usually lose its solubility and form solid precipitates, as the hydrophobic residues from the interior come into contact with water. You can see this phenomenon for yourself if you pour a little bit of vinegar (acetic acid) into milk. The solid clumps that form in the milk are proteins that have come unfolded due to the sudden acidification, and precipitated out of solution.
In recent years, scientists have become increasing interested in the proteins of so-called ‘thermophilic’ (heat-loving) microorganisms that thrive in hot water environments such as geothermal hot springs. While the proteins in most organisms (including humans) will rapidly unfold and precipitate out of solution when put in hot water, the proteins of thermophilic microbes remain completely stable, sometimes even in water that is just below the boiling point. In fact, these proteins typically only gain full biological activity when in appropriately hot water - at room-temperature they act is if they are ‘frozen’. Is the chemical structure of these thermostable proteins somehow unique and exotic? As it turns out, the answer to this question is ‘no’: the overall three-dimensional structures of thermostable proteins look very much like those of ‘normal’ proteins. The critical difference seems to be simply that thermostable proteins have more extensive networks of noncovalent interactions, particularly ion-ion interactions on their surface, that provides them with a greater stability to heat. Interestingly, the proteins of ‘psychrophilic’ (cold-loving) microbes isolated from pockets of water in arctic ice show the opposite characteristic: they have far fewer ion-ion interactions, which gives them greater flexibility in cold temperatures but leads to their rapid unfolding in room temperature water. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/26%3A_Amino_Acids_Peptides_Proteins_and_Nucleic_Acids%3A_Nitrogen-Containing_Polymers_in_Nature/26.04%3A_Peptides__and__Proteins%3A_Amino__Acid__Oligomers_a.txt |
Objectives
After completing this section, you should be able to
1. describe how an Edman degradation is used to determine the sequence of the amino acid residues in peptides containing up to 20 such residues.
2. describe, briefly, how the procedure is modified to deal with peptides and proteins containing more than 20 amino acid residues.
3. write a detailed mechanism for the Edman degradation.
4. determine the structure of a peptide, given a list of the fragments that are produced by a partial acid hydrolysis.
5. determine the structure of a peptide, given a list of the fragments that are produced when the peptide is cleaved by a specific enzyme and the details of the types of bonds cleaved by that enzyme.
6. predict the fragments that would be produced when a peptide of known structure is cleaved by a specific enzyme, given sufficient information about the types of bonds that are cleaved by the enzyme in question.
Key Terms
Make certain that you can define, and use in context, the key term below.
• Edman degradation
Study Notes
The reagent used in the Edman degradation is phenyl isothiocyanate. You may find it helpful to review the relationship between cyanates, isocyanates, thiocyanates and isothiocyanates.
You need not memorize the specific peptide bonds that are broken by the enzymes trypsin and chymotrypsin.
Edman degradation is the process of purifying protein by sequentially removing one residue at a time from the amino end of a peptide. To solve the problem of damaging the protein by hydrolyzing conditions, Pehr Edman created a new way of labeling and cleaving the peptide. Edman thought of a way of removing only one residue at a time, which did not damage the overall sequencing. This was done by adding Phenyl isothiocyanate, which creates a phenylthiocarbamoyl derivative with the N-terminal. The N-terminal is then cleaved under less harsh acidic conditions, creating a cyclic compound of phenylthiohydantoin PTH-amino acid. This does not damage the protein and leaves two constituents of the peptide. This method can be repeated for the rest of the residues, separating one residue at a time.
Edman degradation is very useful because it does not damage the protein. This allows sequencing of the protein to be done in less time. Edman sequencing is done best if the composition of the amino acid is known. As we saw in Section 26.5, to determine the composition of the amino acid, the peptide must be hydrolyzed. This can be done by denaturing the protein and heating it and adding HCl for a long time. This causes the individual amino acids to be separated, and they can be separated by ion exchange chromatography. They are then dyed with ninhydrin and the amount of amino acid can be determined by the amount of optical absorbance. This way, the composition but not the sequence can be determined
Sequencing Larger Proteins
Larger proteins cannot be sequenced by the Edman sequencing because of the less than perfect efficiency of the method. A strategy called divide and conquer successfully cleaves the larger protein into smaller, practical amino acids. This is done by using a certain chemical or enzyme which can cleave the protein at specific amino acid residues. The separated peptides can be isolated by chromatography. Then they can be sequenced using the Edman method, because of their smaller size.
In order to put together all the sequences of the different peptides, a method of overlapping peptides is used. The strategy of divide and conquer followed by Edman sequencing is used again a second time, but using a different enzyme or chemical to cleave it into different residues. This allows two different sets of amino acid sequences of the same protein, but at different points. By comparing these two sequences and examining for any overlap between the two, the sequence can be known for the original protein.
For example, trypsin can be used on the initial peptide to cleave it at the carboxyl side of arginine and lysine residues. Using trypsin to cleave the protein and sequencing them individually with Edman degradation will yield many different individual results. Although the sequence of each individual cleaved amino acid segment is known, the order is scrambled. Chymotrypsin, which cleaves on the carboxyl side of aromatic and other bulky nonpolar residues, can be used. The sequence of these segments overlap with those of the trypsin. They can be overlapped to find the original sequence of the initial protein. However, this method is limited in analyzing larger sized proteins (more than 100 amino acids) because of secondary hydrogen bond interference. Other weak intermolecular bonding such as hydrophobic interactions cannot be properly predicted. Only the linear sequence of a protein can be properly predicted assuming the sequence is small enough. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/26%3A_Amino_Acids_Peptides_Proteins_and_Nucleic_Acids%3A_Nitrogen-Containing_Polymers_in_Nature/26.05%3A_Determination_of_Primary__Structure%3A_Amino__Acid_.txt |
Objectives
After completing this section, you should be able to
1. describe why it is necessary to protect certain amino and carboxyl groups during the synthesis of a peptide.
2. describe, using appropriate equations, how carboxyl groups are protected by ester formation and amino groups are protected by the formation of their tert‑butoxycarbonyl amide derivatives.
3. write a detailed mechanism for the formation of a peptide link between an amino acid with a protected amino group and an amino acid with a protected carboxyl group using dicyclohexylcarbodiimide.
4. outline the five steps required in order to form a dipeptide from two given amino acids.
In order to synthesize a peptide from its component amino acids, two obstacles must be overcome. The first of these is statistical in nature, and is illustrated by considering the dipeptide Ala-Gly as a proposed target. If we ignore the chemistry involved, a mixture of equal molar amounts of alanine and glycine would generate four different dipeptides. These are: Ala-Ala, Gly-Gly, Ala-Gly & Gly-Ala. In the case of tripeptides, the number of possible products from these two amino acids rises to eight. Clearly, some kind of selectivity must be exercised if complex mixtures are to be avoided.
The second difficulty arises from the fact that carboxylic acids and 1º or 2º-amines do not form amide bonds on mixing, but will generally react by proton transfer to give salts (the intermolecular equivalent of zwitterion formation).
From the perspective of an organic chemist, peptide synthesis requires selective acylation of a free amine. To accomplish the desired amide bond formation, we must first deactivate all extraneous amine functions so they do not compete for the acylation reagent. Then we must selectively activate the designated carboxyl function so that it will acylate the one remaining free amine. Fortunately, chemical reactions that permit us to accomplish these selections are well known.
First, the basicity and nucleophilicity of amines are substantially reduced by amide formation. Consequently, the acylation of amino acids by treatment with acyl chlorides or anhydrides at pH > 10, as described earlier, serves to protect their amino groups from further reaction.
Second, acyl halide or anhydride-like activation of a specific carboxyl reactant must occur as a prelude to peptide (amide) bond formation. This is possible, provided competing reactions involving other carboxyl functions that might be present are precluded by preliminary ester formation. Remember, esters are weaker acylating reagents than either anhydrides or acyl halides, as noted earlier.
Finally, dicyclohexylcarbodiimide (DCC) effects the dehydration of a carboxylic acid and amine mixture to the corresponding amide under relatively mild conditions. The structure of this reagent and the mechanism of its action have been described. Its application to peptide synthesis will become apparent in the following discussion.
The strategy for peptide synthesis, as outlined here, should now be apparent. The following example shows a selective synthesis of the dipeptide Ala-Gly.
An important issue remains to be addressed. Since the N-protective group is an amide, removal of this function might require conditions that would also cleave the just formed peptide bond. Furthermore, the harsh conditions often required for amide hydrolysis might cause extensive racemization of the amino acids in the resulting peptide. This problem strikes at the heart of our strategy, so it is important to give careful thought to the design of specific N-protective groups. In particular, three qualities are desired:
1. The protective amide should be easy to attach to amino acids.
2. The protected amino group should not react under peptide forming conditions.
3. The protective amide group should be easy to remove under mild conditions.
A number of protective groups that satisfy these conditions have been devised; and two of the most widely used, carbobenzoxy (Cbz) and t-butoxycarbonyl (BOC or t-BOC), are described here.
The reagents for introducing these N-protective groups are the acyl chlorides or anhydrides shown in the left portion of the above diagram. Reaction with a free amine function of an amino acid occurs rapidly to give the "protected" amino acid derivative shown in the center. This can then be used to form a peptide (amide) bond to a second amino acid. Once the desired peptide bond is created the protective group can be removed under relatively mild non-hydrolytic conditions. Equations showing the protective group removal will be displayed above by are shown above. Cleavage of the reactive benzyl or tert-butyl groups generates a common carbamic acid intermediate (HOCO-NHR) which spontaneously loses carbon dioxide, giving the corresponding amine. If the methyl ester at the C-terminus is left in place, this sequence of reactions may be repeated, using a different N-protected amino acid as the acylating reagent. Removal of the protective groups would then yield a specific tripeptide, determined by the nature of the reactants and order of the reactions. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/26%3A_Amino_Acids_Peptides_Proteins_and_Nucleic_Acids%3A_Nitrogen-Containing_Polymers_in_Nature/26.06%3A_Synthesis_of_Polypeptides%3A_A_Challenge_in_the_App.txt |
Objectives
After completing this section, you should be able to describe, briefly, the Merrifield solid‑phase technique for the synthesis of polypeptides.
Key Terms
Make certain that you can define, and use in context, the key term below:
• solid‑phase method (solid‑phase synthesis)
Study Notes
The solid‑phase used in this method is a polymer support. You will not be examined on the details of the Merrifield solid‑phase method; however, you should be prepared to write a couple of paragraphs describing this important process.
For his work on the synthesis of peptides, Bruce Merrifield was awarded the 1984 Nobel Prize in chemistry.
The synthesis of a peptide of significant length (e.g. ten residues) by this approach requires many steps, and the product must be carefully purified after each step to prevent unwanted cross-reactions. To facilitate the tedious and time consuming purifications, and reduce the material losses that occur in handling, a clever modification of this strategy has been developed. This procedure, known as the Merrifield Synthesis after its inventor R. Bruce Merrifield, involves attaching the C-terminus of the peptide chain to a polymeric solid, usually having the form of very small beads. Separation and purification is simply accomplished by filtering and washing the beads with appropriate solvents. The reagents for the next peptide bond addition are then added, and the purification steps repeated. The entire process can be automated, and peptide synthesis machines based on the Merrifield approach are commercially available. A series of equations illustrating the Merrifield synthesis may be viewed below. The final step, in which the completed peptide is released from the polymer support, is a simple benzyl ester cleavage. This is not shown in the display.
The Merrifield Peptide Synthesis
Two or more moderately sized peptides can be joined together by selective peptide bond formation, provided side-chain functions are protected and do not interfere. In this manner good sized peptides and small proteins may be synthesized in the laboratory. However, even if chemists assemble the primary structure of a natural protein in this or any other fashion, it may not immediately adopt its native secondary, tertiary and quaternary structure. Many factors, such as pH, temperature and inorganic ion concentration influence the conformational coiling of peptide chains. Indeed, scientists are still trying to understand how and why these higher structures are established in living organisms.
26.08: Polypeptides in Nature: Oxygen Transport by t
Oxygen Transport
Many microorganisms and most animals obtain energy by respiration, the oxidation of organic or inorganic molecules by O2. At 25°C, however, the concentration of dissolved oxygen in water in contact with air is only about 0.25 mM. Because of their high surface area-to-volume ratio, aerobic microorganisms can obtain enough oxygen for respiration by passive diffusion of O2 through the cell membrane. As the size of an organism increases, however, its volume increases much more rapidly than its surface area, and the need for oxygen depends on its volume. Consequently, as a multicellular organism grows larger, its need for O2 rapidly outstrips the supply available through diffusion. Unless a transport system is available to provide an adequate supply of oxygen for the interior cells, organisms that contain more than a few cells cannot exist. In addition, O2 is such a powerful oxidant that the oxidation reactions used to obtain metabolic energy must be carefully controlled to avoid releasing so much heat that the water in the cell boils. Consequently, in higher-level organisms, the respiratory apparatus is located in internal compartments called mitochondria, which are the power plants of a cell. Oxygen must therefore be transported not only to a cell but also to the proper compartment within a cell.
Three different chemical solutions to the problem of oxygen transport have developed independently in the course of evolution, as indicated in Table 26.8.13. Mammals, birds, reptiles, fish, and some insects use a heme protein called hemoglobin to transport oxygen from the lungs to the cells, and they use a related protein called myoglobin to temporarily store oxygen in the tissues. Several classes of invertebrates, including marine worms, use an iron-containing protein called hemerythrin to transport oxygen, whereas other classes of invertebrates (arthropods and mollusks) use a copper-containing protein called hemocyanin. Despite the presence of the hem- prefix, hemerythrin and hemocyanin do not contain a metal–porphyrin complex.
Table 26.8.13 Some Properties of the Three Classes of Oxygen-Transport Proteins
Protein Source M per Subunit M per O2 Bound Color (deoxy form) Color (oxy form)
hemoglobin mammals, birds, fish, reptiles, some insects 1 Fe 1 Fe red-purple red
hemerythrin marine worms 2 Fe 2 Fe colorless red
hemocyanin mollusks, crustaceans, spiders 2 Cu 2 Cu colorless blue
Myoglobin and Hemoglobin
Myoglobin is a relatively small protein that contains 150 amino acids. The functional unit of myoglobin is an iron–porphyrin complex that is embedded in the protein (Figure 26.8.1). In myoglobin, the heme iron is five-coordinate, with only a single histidine imidazole ligand from the protein (called the proximal histidine because it is near the iron) in addition to the four nitrogen atoms of the porphyrin. A second histidine imidazole (the distal histidine because it is more distant from the iron) is located on the other side of the heme group, too far from the iron to be bonded to it. Consequently, the iron atom has a vacant coordination site, which is where O2 binds.
Figure 26.8.1: The Structure of Deoxymyoglobin, Showing the Heme Group. The iron in deoxymyoglobin is five-coordinate, with one histidine imidazole ligand from the protein. Oxygen binds at the vacant site on iron.
In the ferrous form (deoxymyoglobin), the iron is five-coordinate and high spin. Because high-spin Fe2+ is too large to fit into the “hole” in the center of the porphyrin, it is about 60 pm above the plane of the porphyrin. When O2 binds to deoxymyoglobin to form oxymyoglobin, the iron is converted from five-coordinate (high spin) to six-coordinate (low spin; Figure 26.8.2). Because low-spin Fe2+ and Fe3+ are smaller than high-spin Fe2+, the iron atom moves into the plane of the porphyrin ring to form an octahedral complex. The O2 pressure at which half of the molecules in a solution of myoglobin are bound to O2 (P1/2) is about 1 mm Hg (1.3 × 10−3 atm).
affinity unchanged, which is important because carbon monoxide is produced continuously in the body by degradation of the porphyrin ligand (even in nonsmokers). Under normal conditions, CO occupies approximately 1% of the heme sites in hemoglobin and myoglobin. If the affinity of hemoglobin and myoglobin for CO were 100 times greater (due to the absence of the distal histidine), essentially 100% of the heme sites would be occupied by CO, and no oxygen could be transported to the tissues. Severe carbon-monoxide poisoning, which is frequently fatal, has exactly the same effect. Thus the primary function of the distal histidine appears to be to decrease the CO affinity of hemoglobin and myoglobin to avoid self-poisoning by CO. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/26%3A_Amino_Acids_Peptides_Proteins_and_Nucleic_Acids%3A_Nitrogen-Containing_Polymers_in_Nature/26.07%3A_Merrifield__Solid-Phase_Peptide__Synthesis.txt |
Objectives
After completing this section, you should be able to
• Describe briefly how DNA sequencing is carried out.
DNA sequencing determines the order of nucleotide bases within a given fragment of DNA. This information can be used to infer the RNA or protein sequence encoded by the gene, from which further inferences may be made about the gene’s function and its relationship to other genes and gene products. DNA sequence information is also useful in studying the regulation of gene expression. If DNA sequencing is applied to the study of many genes, or even a whole genome, it is considered an example of genomics.
While techniques to sequence proteins have been around since the 1950s, techniques to sequence DNA were not developed until the mid-1970s, when two distinct sequencing methods were developed almost simultaneously, one by Walter Gilbert’s group at Harvard University, the other by Frederick Sanger’s group at Cambridge University. However, until the 1990s, the sequencing of DNA was a relatively expensive and long process. Using radiolabeled nucleotides also compounded the problem through safety concerns. With currently-available technology and automated machines, the process is cheaper, safer, and can be completed in a matter of hours. The Sanger sequencing method was used for the human genome sequencing project, which was finished its sequencing phase in 2003, but today both it and the Gilbert method have been largely replaced by better methods.
Restriction Enzymes
To be able to sequence DNA, it is first necessary to cut it into smaller fragments. What is needed is a way to cleave the DNA molecule at a few precisely-located sites so that a small set of homogeneous fragments are produced. To cut DNA at known locations, researchers use restriction endonucleases enzymes that have been purified from various bacterial species, and which can be purchased from various commercial sources. REs occur naturally in bacteria, where they specifically recognize short stretches of nucleotides in DNA and catalyze double-strand breaks at or near the recognition site (also known as a restriction site). These enzymes are usually named after the bacterium from which they were first isolated. For example, EcoRI and EcoRV are both enzymes from E. coli.
Restriction enzymes like EcoRI are frequently called 6-cutters, because they recognize a 6-nucleotide sequence. Assuming a random distribution of A, C, G and Ts in DNA, probability predicts that a recognition site for a 6-cutter should occur about once for every 4096 bp (46) in DNA. Of course, the distribution of nucleotides in DNA is not random, so the actual sizes of DNA fragments produced by EcoRI range from hundreds to many thousands of base pairs, but the mean size is close to 4000 bp. DNA fragments of this length are useful in the lab, since they long enough to contain the coding sequence for proteins and are well-resolved on agarose gels.
EcoRI recognizes the sequence G A A T T C in double stranded DNA. This recognition sequence is a palindrome with a two-fold axis of symmetry, because reading from 5’ to 3’ on either strand of the helix gives the same sequence. The palindromic nature of the restriction site is more obvious in the figure below. The dot in the center of the restriction site denotes the axis of symmetry. EcoRI catalyzes the hydrolysis of the phosphodiester bonds between G and A on both DNA strands. The restriction fragments generated in the reaction have short single-stranded tails at the 5’-ends. These ends are often referred to as “sticky ends,” because of their ability to form hydrogen bonds with complementary DNA sequences.
Reading DNA Sequences
We will discuss one method of reading the sequence of DNA. This method, developed by Sanger won him a second Nobel prize. Sanger sequencing, also known as chain-termination sequencing, requires a single-stranded DNA template, a DNA primer, a DNA polymerase, normal deoxynucleotidetriphosphates (dNTPs), and modified nucleotides (dideoxyNTPs - ddNTP) that terminate DNA strand elongation. These chain-terminating nucleotides lack a 3′-OH group required for the formation of a phosphodiester bond between two nucleotides, causing DNA polymerase to cease extension of DNA when a ddNTP is incorporated.
Four reaction tubes are set up, each containing the template DNA to be sequenced, a primer of known sequence, all four of the standard deoxynucleotides (dATP, dGTP, dCTP and dTTP), and the DNA polymerase. To each reaction is added only one of the four dideoxynucleotides (ddATP, ddGTP, ddCTP, or ddTTP) which has been fluorescently labeled. Most of the time in a Sanger sequencing reaction, DNA Polymerase will add a proper dNTP to the growing strand it is synthesizing in vitro. But at random locations, it will instead add a ddNTP. When it does, that strand will be terminated at the ddNTP just added. If enough template DNAs are included in the reaction mix, each one will have the labeled ddNTP inserted at a different random location, and there will be at least one DNA terminated at each different nucleotide along its length for as long as the in vitro reaction can take place (about 900 nucleotides under optimal conditions.)
After the reactions are over, the newly synthesized strands can be denatured from the template, and then separated by capillary electrophoresis or other equivalent methods. Since each band differs in length by one nucleotide, and the identity of that nucleotide is known from its fluorescence, the DNA sequence can be read simply from the order of the colors in successive bands.
As each differently-sized fragment exits the capillary column, a laser excites the florescent tag on its terminal nucleotide. From the color of the resulting florescence, a computer can keep track of which nucleotide was present as the terminating nucleotide. The computer also keeps track of the order in which the terminating nucleotides appeared, which is the sequence of the DNA used in the original reaction. In practice, the maximum length of sequence that can be read from a single sequencing reaction is about 700 bp.
Scientists now know the sequence of all the 3 billion DNA base pairs in the entire human genome. This knowledge was attained by the Human Genome Project (HGP), a \$3 billion, international scientific research project that was formally launched in 1990. The project was completed in 2003, two years ahead of its 15-year projected deadline. Determining the sequence of the billions of base pairs that make up human DNA was the main goal of the HGP. Another goal was mapping the location and determining the function of all the genes in the human genome. There are only about 20,500 genes in human beings. If modern methods were used it might bring the cost of sequencing the human genome down from the initial billion dollar range to \$100.
Example 28.6.1
You will pretend to sequence a single stranded piece of DNA as shown below. The new nucleotides are added by the enzyme DNA polymerase to the primer, GACT, in the 5' to 3' direction. You will set up 4 reaction tubes, Each tube contains all the dXTP's. In addition, add ddATP to tube 1, ddTTP to tube 2, ddCTP to tube 3, and ddGTP to tube 4. For each separate reaction mixture, determine all the possible sequences made by writing the possible sequences on one of the unfinished complementary sequences below. Cut the completed sequences from the page, determine the size of the polynucleotide sequences made, and place them as they would migrate (based on size) in the appropriate lane of a imaginary gel which you have drawn on a piece of paper. Lane 1 will contain the nucleotides made in tube 1, etc. Then draw lines under the positions of the cutout nucleotides to represent DNA bands in the gel. Read the sequence of the complementary DNA synthesized. Then write the sequence of the ssDNA that was to be sequenced.
• 5' T C A A C G A T C T G A 3' (STAND TO SEQUENCE)
• 3' G A C T 5' (primer)
• 3' G A C T 5' (primer)
• 3' G A C T 5' (primer)
• 3' G A C T 5' (primer)
• 3' G A C T 5' (primer)
• 3' G A C T 5' (primer)
• 3' G A C T 5' (primer)
• 3' G A C T 5' (primer)
Since the DNA fragments have no detectable color, they can not be directly visualized in the gel. Alternative methods are used. In the one described above, radiolabeled ddXTP's where used. Once the sequencing gel is run, it can be dried and the bands visualized by radioautography (also called autoradiography). A place of x-ray film is placed over the dried gel in a dark environment. The radiolabeled bands will emit radiation which will expose the x-ray film directly over the bands. The film can be developed to detect the bands. In a newer technique, the primer can be labeled with a flourescent dye. If a different dye is used for each reaction mixture, all the reaction mixtures can be run in one lane of a gel. (Actually only one reaction mix containing all the ddXTP's together need be performed.) The gel can then be scanned by a laser, which detects fluorescence from the dyes, each at a different wavelength. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/26%3A_Amino_Acids_Peptides_Proteins_and_Nucleic_Acids%3A_Nitrogen-Containing_Polymers_in_Nature/26.11%3A_DNA__Sequencing_and_Synthesis%3A_Cornerstones_of_Ge.txt |
Learning Objectives
After reading this chapter and completing ALL the exercises, a student can be able to
• discuss the origins of organic chemistry - refer to section 1.1
• use and apply the language of Atomic Structure (atomic number, mass number, isotopes) - refer to section 1.2
• draw, interpret, and convert between Lewis (Kekule), Condensed, and Bond-line Structures - refer to sections 1.3, 1.4, 1.5, and 1.6
• apply bonding patterns and polarity to organic compounds - refer to section 1.7 and 1.8
• identify polar bonds and compounds - refer to section 1.9
• draw resonance forms and predict the relative contribution of each resonance form to the overall structure of the compound or ion - refer to section 1.10
• recognize acids and bases - refer to sections 1.11 and 1.12
• use the definition of Lewis Acids and Bases to recognize electron movement in reactions - refer to section 1.13
• predict reaction products of acid-base reactions - refer to sections 1.11, 1.12 ,and 1.13
• determine relative strengths of acids and bases from their pKa values - refer to section 1.14
• determine the form of an acid or base at a specified pH (given the pKa) - refer to section 1.14
• predict relative strengths of acids and bases from their structure, bonding and resonance - refer to section 1.15
• determine the empirical and molecular formulas from combustion data - refer to section 1.16
01: Introduction and Review
Learning Objective
• discuss the origins of organic chemistry
All living things on earth are formed mostly of carbon compounds. The prevalence of carbon compounds in living things has led to the epithet “carbon-based” life. The truth is we know of no other kind of life. Early chemists regarded substances isolated from organisms (plants and animals) as a different type of matter that could not be synthesized artificially, and these substances were thus known as organic compounds. The widespread belief called vitalism held that organic compounds were formed by a vital force present only in living organisms. The German chemist Friedrich Wöhler was one of the early chemists to refute this aspect of vitalism, when, in 1828, he reported the synthesis of urea, a component of many body fluids, from nonliving materials. Since then, it has been recognized that organic molecules obey the same natural laws as inorganic substances, and the category of organic compounds has evolved to include both natural and synthetic compounds that contain carbon. Some carbon-containing compounds are not classified as organic, for example, carbonates and cyanides, and simple oxides, such as \(\ce{CO}\) and \(\ce{CO2}\). Although a single, precise definition has yet to be identified by the chemistry community, most agree that a defining trait of organic molecules is the presence of carbon as the principal element, bonded to hydrogen and other carbon atoms.
Today, organic compounds are key components of plastics, soaps, perfumes, sweeteners, fabrics, pharmaceuticals, and many other substances that we use every day. The value to us of organic compounds ensures that organic chemistry is an important discipline within the general field of chemistry. In this chapter, we discuss why the element carbon gives rise to a vast number and variety of compounds, how those compounds are classified, and the role of organic compounds in representative biological and industrial settings. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/01%3A_Introduction_and_Review/1.01%3A_The_Origins_of_Organic_Chemistry.txt |
Learning Objective
• Use and apply the language of Atomic Structure (atomic number, mass number, isotopes)
The precise physical nature of atoms finally emerged from a series of elegant experiments carried out between 1895 and 1915. The most notable of these achievements was Ernest Rutherford's famous 1911 alpha-ray scattering experiment, which established that
• Almost all of the mass of an atom is contained within a tiny (and therefore extremely dense) nucleus which carries a positive electric charge whose value identifies each element and is known as the atomic number of the element.
• Almost all of the volume of an atom consists of empty space in which electrons, the fundamental carriers of negative electric charge, reside. The extremely small mass of the electron (1/1840 the mass of the hydrogen nucleus) causes it to behave as a quantum particle, which means that its location at any moment cannot be specified; the best we can do is describe its behavior in terms of the probability of its manifesting itself at any point in space. It is common (but somewhat misleading) to describe the volume of space in which the electrons of an atom have a significant probability of being found as the electron cloud. The latter has no definite outer boundary, so neither does the atom. The radius of an atom must be defined arbitrarily, such as the boundary in which the electron can be found with 95% probability. Atomic radii are typically 30-300 pm.
The nucleus is itself composed of two kinds of particles. Protons are the carriers of positive electric charge in the nucleus; the proton charge is exactly the same as the electron charge, but of opposite sign. This means that in any [electrically neutral] atom, the number of protons in the nucleus (often referred to as the nuclear charge) is balanced by the same number of electrons outside the nucleus. The other nuclear particle is the neutron. As its name implies, this particle carries no electrical charge. Its mass is almost the same as that of the proton. Most nuclei contain roughly equal numbers of neutrons and protons, so we can say that these two particles together account for almost all the mass of the atom.
Because the electrons of an atom are in contact with the outside world, it is possible for one or more electrons to be lost, or some new ones to be added. The resulting electrically-charged atom is called an ion.
Elements
To date, about 115 different elements have been discovered; by definition, each is chemically unique. To understand why they are unique, you need to understand the structure of the atom (the fundamental, individual particle of an element) and the characteristics of its components. Atoms consist of electrons, protons, and neutrons. Although this is an oversimplification that ignores the other subatomic particles that have been discovered, it is sufficient for discussion of chemical principles. Some properties of these subatomic particles are summarized in Table $1$, which illustrates three important points:
1. Electrons and protons have electrical charges that are identical in magnitude but opposite in sign. Relative charges of −1 and +1 are assigned to the electron and proton, respectively.
2. Neutrons have approximately the same mass as protons but no charge. They are electrically neutral.
3. The mass of a proton or a neutron is about 1836 times greater than the mass of an electron. Protons and neutrons constitute the bulk of the mass of atoms.
The discovery of the electron and the proton was crucial to the development of the modern model of the atom and provides an excellent case study in the application of the scientific method. In fact, the elucidation of the atom’s structure is one of the greatest detective stories in the history of science.
Table $1$: Properties of Subatomic Particles*
Particle Mass (g) Atomic Mass (amu) Electrical Charge (coulombs) Relative Charge
electron $9.109 \times 10^{-28}$ 0.0005486 −1.602 × 10−19 −1
proton $1.673 \times 10^{-24}$ 1.007276 +1.602 × 10−19 +1
neutron $1.675 \times 10^{-24}$ 1.008665 0 0
In most cases, the symbols for the elements are derived directly from each element’s name, such as C for carbon, U for uranium, Ca for calcium, and Po for polonium. Elements have also been named for their properties [such as radium (Ra) for its radioactivity], for the native country of the scientist(s) who discovered them [polonium (Po) for Poland], for eminent scientists [curium (Cm) for the Curies], for gods and goddesses [selenium (Se) for the Greek goddess of the moon, Selene], and for other poetic or historical reasons. Some of the symbols used for elements that have been known since antiquity are derived from historical names that are no longer in use; only the symbols remain to indicate their origin. Examples are Fe for iron, from the Latin ferrum; Na for sodium, from the Latin natrium; and W for tungsten, from the German wolfram. Examples are in Table $2$.
Table $2$: Element Symbols Based on Names No Longer in Use
Element Symbol Derivation Meaning
antimony Sb stibium Latin for “mark”
copper Cu cuprum from Cyprium, Latin name for the island of Cyprus, the major source of copper ore in the Roman Empire
gold Au aurum Latin for “gold”
iron Fe ferrum Latin for “iron”
lead Pb plumbum Latin for “heavy”
mercury Hg hydrargyrum Latin for “liquid silver”
potassium K kalium from the Arabic al-qili, “alkali”
silver Ag argentum Latin for “silver”
sodium Na natrium Latin for “sodium”
tin Sn stannum Latin for “tin”
tungsten W wolfram German for “wolf stone” because it interfered with the smelting of tin and was thought to devour the tin
Recall that the nuclei of most atoms contain neutrons as well as protons. Unlike protons, the number of neutrons is not absolutely fixed for most elements. Atoms that have the same number of protons, and hence the same atomic number, but different numbers of neutrons are called isotopes. All isotopes of an element have the same number of protons and electrons, which means they exhibit the same chemistry. The isotopes of an element differ only in their atomic mass, which is given by the mass number (A), the sum of the numbers of protons and neutrons.
The element carbon (C) has an atomic number of 6, which means that all neutral carbon atoms contain 6 protons and 6 electrons. In a typical sample of carbon-containing material, 98.89% of the carbon atoms also contain 6 neutrons, so each has a mass number of 12. An isotope of any element can be uniquely represented as $^A_Z X$, where X is the atomic symbol of the element. The isotope of carbon that has 6 neutrons is therefore $_6^{12} C$. The subscript indicating the atomic number is actually redundant because the atomic symbol already uniquely specifies Z. Consequently, $_6^{12} C$ is more often written as 12C, which is read as “carbon-12.” Nevertheless, the value of Z is commonly included in the notation for nuclear reactions because these reactions involve changes in Z.
Figure $2$: Formalism used for identifying specific nuclide (any particular kind of nucleus)
In addition to $^{12}C$, a typical sample of carbon contains 1.11% $_6^{13} C$ (13C), with 7 neutrons and 6 protons, and a trace of $_6^{14} C$ (14C), with 8 neutrons and 6 protons. The nucleus of 14C is not stable, however, but undergoes a slow radioactive decay that is the basis of the carbon-14 dating technique used in archaeology. Many elements other than carbon have more than one stable isotope; tin, for example, has 10 isotopes. The properties of some common isotopes are in Table $3$.
Table $3$: Properties of Selected Isotopes
Element Symbol Atomic Mass (amu) Isotope Mass Number Isotope Masses (amu) Percent Abundances (%)
hydrogen H 1.0079 1 1.007825 99.9855
2 2.014102 0.0115
boron B 10.81 10 10.012937 19.91
11 11.009305 80.09
carbon C 12.011 12 12 (defined) 99.89
13 13.003355 1.11
oxygen O 15.9994 16 15.994915 99.757
17 16.999132 0.0378
18 17.999161 0.205
iron Fe 55.845 54 53.939611 5.82
56 55.934938 91.66
57 56.935394 2.19
58 57.933276 0.33
uranium U 238.03 234 234.040952 0.0054
235 235.043930 0.7204
238 238.050788 99.274
Sources of isotope data: G. Audi et al., Nuclear Physics A 729 (2003): 337–676; J. C. Kotz and K. F. Purcell, Chemistry and Chemical Reactivity, 2nd ed., 1991.
Example $1$
An element with three stable isotopes has 82 protons. The separate isotopes contain 124, 125, and 126 neutrons. Identify the element and write symbols for the isotopes.
Given: number of protons and neutrons
Asked for: element and atomic symbol
Strategy:
1. Refer to the periodic table and use the number of protons to identify the element.
2. Calculate the mass number of each isotope by adding together the numbers of protons and neutrons.
3. Give the symbol of each isotope with the mass number as the superscript and the number of protons as the subscript, both written to the left of the symbol of the element.
Solution:
A The element with 82 protons (atomic number of 82) is lead: Pb.
B For the first isotope, A = 82 protons + 124 neutrons = 206. Similarly, A = 82 + 125 = 207 and A = 82 + 126 = 208 for the second and third isotopes, respectively. The symbols for these isotopes are $^{206}_{82}Pb$, $^{207}_{82}Pb$, and $^{208}_{82}Pb$, which are usually abbreviated as $^{206}Pb$, $^{207}Pb$, and $^{208}Pb$.
Exercise $1$
Identify the element with 35 protons and write the symbols for its isotopes with 44 and 46 neutrons.
Answer
$\ce{^{79}_{35}Br}$ and $\ce{^{81}_{35}Br}$ or, more commonly, $\ce{^{79}Br}$ and $\ce{^{81}Br}$.
Summary
The atom consists of discrete particles that govern its chemical and physical behavior. Each atom of an element contains the same number of protons, which is the atomic number (Z). Neutral atoms have the same number of electrons and protons. Atoms of an element that contain different numbers of neutrons are called isotopes. Each isotope of a given element has the same atomic number but a different mass number (A), which is the sum of the numbers of protons and neutrons. The relative masses of atoms are reported using the atomic mass unit (amu), which is defined as one-twelfth of the mass of one atom of carbon-12, with 6 protons, 6 neutrons, and 6 electrons. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/01%3A_Introduction_and_Review/1.02%3A_Principles_of_Atomic_Structure_%28Review%29.txt |
Learning Objective
Draw, interpret, and convert between Lewis (Kekule), Condensed, and Bond-line Structures
Note: The review of general chemistry in sections 1.3 - 1.6 is integrated into the above Learning Objective for organic chemistry in sections 1.7 and 1.8.
The primary skills needed are the ability to determine the electron configurations of elements following the concepts of "aufbau" or "build up", Hund's Rule, and the Pauli Exclusion Principle, as well as visualizing the orbitals, subshells, and shells spatially and energetically. Converting electron configurations to orbital diagrams is a useful skill as we transition to organic chemistry. These skills are practiced in the next section.
The Wave Nature of Light
A wave is a periodic oscillation by which energy is transmitted through space. All waves are periodic, repeating regularly in both space and time. Waves are characterized by several interrelated properties.
• Electronic structure: arrangement of electrons in atoms
• Electromagnetic radiation: aka radiant energy; form of energy that has wave characteristics and carries energy through space. All types of electromagnetic radiation move through a vacuum at a speed of 3.00 X 108 m/s (speed of light).
• Wavelength: the distance between identical points on successive waves
• Frequency: the number of complete wavelengths that pass a given point in 1s
$\nu λ = c$
where $\nu$ = frequency, $λ$= wavelength, and $c$ = speed of light
Wavelength is expressed in units of length.
• Electromagnetic spectrum: various types of electromagnetic radiations arranged in order of increasing wavelength.
Frequency is expressed in Hertz (Hz), also denoted by s-1 or /s
Quantum Effects and Photons
• Quantum: smallest quantity of energy that can be emitted or absorbed as electromagnetic radiation
$E = hv$
where $E$ = energy, $h$ = Planck’s constant, $\nu$ = frequency
Planck’s constant = 6.63 X 10-34 J/s
According to Planck’s theory, energy is always emitted or absorbed in whole-number multiples of hv, for example, hv, 2hv, 3hv, and so forth. We say that the allowed energies are quantized (that is, their values are restricted to certain quantities).
Quantized Energy and Photons
Blackbody radiation is the radiation emitted by hot objects and could not be explained with classical physics. Max Planck postulated that energy was quantized and may be emitted or absorbed only in integral multiples of a small unit of energy, known as a quantum. The energy of a quantum is proportional to the frequency of the radiation; the proportionality constant h is a fundamental constant (Planck’s constant). Albert Einstein used the quantization of energy to explain the photoelectric effect
• The photoelectric effect: when photons of sufficiently high energy strike a metal surface, electrons are emitted from the metal. The emitted electrons are drawn toward the other electrode, which is a positive terminal. As a result, current flows in a circuit.
• Photon: smallest increment (a quantum) of radiant energy; a photon of light with frequency v has an energy equal to hv.
• When a photon strikes the metal, its energy is transferred to an electron in the metal. A certain amount of energy is required for the electron to overcome the attractive forces that hold it within the metal. If the photons have less energy that this energy threshold, the electrons cannot escape from the metal surface. If a photon has sufficient energy, an electron is emitted. If the photon has more energy than necessary, the excess appears as kinetic energy of the emitted electron.
Line Spectra and the Bohr Model
There is an intimate connection between the atomic structure of an atom and its spectral characteristics. Most light is polychromatic and contains light of many wavelengths. Light that has only a single wavelength is monochromatic and is produced by devices called lasers, which use transitions between two atomic energy levels to produce light in a very narrow range of wavelengths. Atoms can also absorb light of certain energies, resulting in a transition from the ground state or a lower-energy e
Line Spectra
Radiation composed of a single wavelength is said to be monochromatic.
• Spectrum: distribution among various wavelengths of the radiant energy emitted or absorbed by an object
• Continuous spectrum: rainbow of colors, containing light of all wavelengths. Not all radiation sources produce a continuous spectrum
• Line spectrum: spectrum containing radiation of only specific wavelengths
$v = C \left(\dfrac{1}{2^2} – \dfrac{1}{n^2}\right)$
with $n$ = 3, 4, 5, 6, and $C = 3.29 \times 10^{15} s^{-1}$ (constant)
Bohr’s Model
Electrons in a permitted orbit have a specific energy and are said to be in an "allowed" energy state. An electron in an allowed energy state will not radiate energy and therefore will not spiral into the nucleus.
$E_n = -R_H \dfrac{1}{n^2}$
• $R_H$ = Rydberg constant: 2.18 X 10-18 J
• $n$ = principal quantum number, corresponds to the different allowed orbits for the electron
All energies given by this equation will be negative. The lower (more negative) the energy is, the more stable the atom is. The lowest energy state is that for which n=1.
• Ground state: lowest energy state of an atom, $n=1$
• Excited state: when the electron is in higher energy orbit (less negative), n=2 or higher
If n becomes infinitely large (∞), the electron is completely separated from the nucleus:
$E_∞ = (-2.18 \times 10^{-18} J) \left(\dfrac{1}{∞^2}\right) = 0$
Thus, the state in which the electron is removed from the nucleus is the reference, or zero-energy, state of the hydrogen atom. It is important to remember that this zero-energy state is higher in energy than the states with negative energies
Electrons can change from one energy state to another by absorbing or emitting radiant energy. Radiant energy must be absorbed for an electron to move to a higher energy state, but is emitted when the electron moves to a lower energy state. .
$\Delta E = E_f – E_i$
• If $n_f > n_i$, then ∆E is positive, radiant energy is absorbed
• If $n_f < n_i$, then ∆E is negative, radiant energy is emitted
The Wave Behavior of Matter
An electron possesses both particle and wave properties. Louis de Broglie showed that the wavelength of a particle is equal to Planck’s constant divided by the mass times the velocity of the particle. The electron in Bohr’s circular orbits could thus be described as a standing wave, one that does not move through space. Werner Heisenberg’s uncertainty principle states that it is impossible to precisely describe both the location and the speed of particles that exhibit wavelike behavior.
• Momentum: the product of the mass, m, and the velocity, v, of a particle
• Matter waves: term used to describe the wave characteristics of a particle
$λ = \dfrac{h}{mv}$
where $λ$ is the wavelength, $h$ is Planck’s constant, $m$ is the particle mass, and $v$ is the velocity
The Uncertainty Principle
• Uncertainty principle: theory first put forth by Heisenberg, states that is it impossible to determine both the exact momentum of the electron and its exact location.
Quantum Mechanics and Atomic Orbitals
There is a relationship between the motions of electrons in atoms and molecules and their energies that is described by quantum mechanics. Because of wave–particle duality, scientists must deal with the probability of an electron being at a particular point in space. To do so required the development of quantum mechanics, which uses wavefunctions to describe the mathematical relationship between the motion of electrons in atoms and molecules and their energies.
• Wave functions: represented by ψ, square of wave function, ψ2, provides information about an electron’s location when it is in an allowed energy state.
• Probability density: represented by ψ2, value that represents the probability that an electron will be found at a given point in space
• Electron density: the probability of finding and electron at any particular point in an atom. Equals ψ2.
Orbitals and Quantum Numbers
-Orbital: allowed energy state of an electron in the quantum-mechanical model of the atom; also used to describe the spatial distribution of an electron. Defined by the value of 3 quantum numbers; n, l, and ml.
Value of l
0
1
2
3
Letter used
s
p
d
f
1. The principal quantum number, n, can have integral values of 1, 2, 3 and so forth. As n increases, the orbital becomes larger; the electron has a higher energy and is farther away from the nucleus.
2. The second quantum number, l, can have integral values from 0 to n – 1 for each value of n. This quantum number defines the shape of the orbital. Generally designated by the letters s, p, d, and f. These correspond to values ranging from 0 to 3.
3. The magnetic quantum number, ml, can have integral values between l and –l, including zero. This quantum number describes the orientation of the orbital in space.
Electron shell: collection of orbitals with the same value of n
Subshell: one or more orbitals with the same set of n and l values
1. Each shell is divided into the number of subshells equal to the principal quantum number, n, for that shell. The first shell consists of only the 1s subshell; the second shell consists of two subshells, 2s and 2p; the third of three subshell, 3s, 3p and 3d, and so forth.
2. Each subshell is divided into orbitals. Each s subshell consists of one orbital; each p subshell of three orbitals, each d subshell of five, and each f subshell of seven orbitals.
3D Representation of Orbitals
Orbitals with l = 0 are s orbitals and are spherically symmetrical, with the greatest probability of finding the electron occurring at the nucleus. Orbitals with values of n > 1 and l = 0 contain one or more nodes. Orbitals with l = 1 are p orbitals and contain a nodal plane that includes the nucleus, giving rise to a dumbbell shape. Orbitals with l = 2 are d orbitals and have more complex shapes with at least two nodal surfaces. l = 3 orbitals are f orbitals, which are still more complex.
The s Orbitals: 1s orbital: most stable, spherically symmetric, figure indicates that the probability decreases as we move away from the nucleus. ALL s ORBITALS ARE SPHERICALLY SYMMETRIC.
• Nodal surfaces (nodes): intermediate regions where ψ2 goes to zero. The number of nodes increases with increasing value for the principal quantum number, n.
The p Orbitals: Electron density is concentrated on two sides of the nucleus, separated by a node at the nucleus. The orbitals of a given subshell have the same size and shape but differ from each other in orientation. The axis along which the orbital is oriented is not related to ml.
Many-Electron Atoms
In addition to the three quantum numbers ($n$, $l$, $m_l$) dictated by quantum mechanics, a fourth quantum number is required to explain certain properties of atoms. This is the electron spin quantum number ($m_s$), which can have values of +½ or −½ for any electron, corresponding to the two possible orientations of an electron in a magnetic field. This is important for chemistry because the Pauli exclusion principle implies that no orbital can contain more than two electrons (with opposite spin).
Although the shapes of the orbitals for many-electron atoms are the same as those for hydrogen, the presence of more than one electron greatly changes the energies of the orbitals. In hydrogen, the energy of an orbital depends only on its principal quantum number, however, in many-electron atoms, electron-electron repulsions cause different subshells to be at different energies
Effective Nuclear Charge
• Effective nuclear charge: net positive charge attracting electrons
$Z_{eff} = Z – S$
where $Z_{eff}$ is the effective nuclear charge, $Z$ is the number of protons in the nucleus, and $S$ is the average number of electrons between the nucleus and electron in question.
• Screening effect: effect of inner electrons in decreasing the nuclear charge experienced by outer electrons
Energies of Orbitals
The extent to which an electron will be screened by the other electrons depends on its electron distribution as we move outward from the nucleus.
• In a many-electron atom, for a given value of n, Zeff decreases with increasing value of l.
• In a many-electron atom, for a given value of n, the energy of an orbital increases with increasing value of l.
Degenerate: orbitals that have the same energy
Electron Spin and the Pauli Exclusion Principle
• Electron spin: property of the electron that makes it behave as though it were a tiny magnet. The electron behaves as if it were spinning on its axis; electron spin is quantized.
• Electron spin quantum number: denoted as ms. It can only have two possible values, +½ and –½, which we can interpret as indicating the two opposite directions in which the electron can spin.
• Pauli exclusion principle: states that no two electrons in an atom can have the same set of four quantum numbers n, l, ml, ms. This means that if we wish to put two electrons in an orbital and satisfy Pauli’s exclusion principle, our only choice is to assign different ms values to the electrons. Because there are only two values, we can conclude that an orbital can hold a maximum of two electrons and they must have opposite spins.
Electron Configurations
Based on the Pauli principle and a knowledge of orbital energies obtained using hydrogen-like orbitals, it is possible to construct the periodic table by filling up the available orbitals beginning with the lowest-energy orbitals (the aufbau principle), which gives rise to a particular arrangement of electrons for each element (its electron configuration). Hund’s rule says that the lowest-energy arrangement of electrons is the one that places them in degenerate orbitals with parallel spins.
• Electron configuration: the way in which the electrons are distributed among the various orbitals. The most stable, or ground, electron configuration of an atom is that in which the electrons are in the lowest possible energy level
• Orbital diagram: representation of electron configuration in which each orbital is represented by a box and each electron by a half-arrow. A half-arrow pointing upward represents an electron with positive spin; one pointing downward represents an electron with a negative spin.
Writing Electron Configurations
• Hund’s rule: rule stating that electrons occupy degenerate orbitals in such a way as to maximize the number of electrons with the same spin. In other words, each orbital has one electron placed in it before paring of electron in orbitals occurs. Note that this rule applies to orbitals that are degenerate, which means that they have the same energy.
• Valence electrons: electrons in the outer shells
• Core electrons: electrons in the inner shells
• Transition elements: aka Transition metals; elements of the d orbitals
• Lanthanide elements: aka Rare-earth elements; 14 elements of the 4f orbitals, # 58-71
• Actinide elements: 14 elements of 5f orbitals, # 90-103. Most are not found in nature.
Electron Configurations and the Periodic Table
The arrangement of atoms in the periodic table results in blocks corresponding to filling of the ns, np, nd, and nf orbitals to produce the distinctive chemical properties of the elements in the s block, p block, d block, and f block, respectively.
• Main group elements: aka Representatives; s and p block elements
• F-block metals: 28 elements located below the table, f block elements | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/01%3A_Introduction_and_Review/1.03%3A_Electronic_Structure_%28Review%29.txt |
Learning Objective
Draw, interpret, and convert between Lewis (Kekule), Condensed, and Bond-line Structures
Note: The review of general chemistry in sections 1.3 - 1.6 is integrated into the above Learning Objective for organic chemistry in sections 1.7 and 1.8.
The electron configuration of an atom is the representation of the arrangement of electrons distributed among the orbital shells and subshells. Commonly, the electron configuration is used to describe the orbitals of an atom in its ground state, but it can also be used to represent an atom that has ionized into a cation or anion by compensating with the loss of or gain of electrons in their subsequent orbitals. Many of the physical and chemical properties of elements can be correlated to their unique electron configurations. The valence electrons, electrons in the outermost shell, are the determining factor for the unique chemistry of the element.
Introduction
Before assigning the electrons of an atom into orbitals, one must become familiar with the basic concepts of electron configurations. Every element on the Periodic Table consists of atoms, which are composed of protons, neutrons, and electrons. Electrons exhibit a negative charge and are found around the nucleus of the atom in electron orbitals, defined as the volume of space in which the electron can be found within 95% probability. The four different types of orbitals (s,p,d, and f) have different shapes, and one orbital can hold a maximum of two electrons. The p, d, and f orbitals have different sublevels, thus can hold more electrons.
As stated, the electron configuration of each element is unique to its position on the periodic table. The energy level is determined by the period and the number of electrons is given by the atomic number of the element. Orbitals on different energy levels are similar to each other, but they occupy different areas in space. The 1s orbital and 2s orbital both have the characteristics of an s orbital (radial nodes, spherical volume probabilities, can only hold two electrons, etc.) but, as they are found in different energy levels, they occupy different spaces around the nucleus. Each orbital can be represented by specific blocks on the periodic table. The s-block is the region of the alkali metals including helium (Groups 1 & 2), the d-block are the transition metals (Groups 3 to 12), the p-block are the main group elements from Groups 13 to 18, and the f-block are the lanthanides and actinides series.
Using the periodic table to determine the electron configurations of atoms is key, but also keep in mind that there are certain rules to follow when assigning electrons to different orbitals. The periodic table is an incredibly helpful tool in writing electron configurations. For more information on how electron configurations and the periodic table are linked, visit the Connecting Electrons to the Periodic Table module.
Rules for Assigning Electron Orbitals
Occupation of Orbitals
Electrons fill orbitals in a way to minimize the energy of the atom. Therefore, the electrons in an atom fill the principal energy levels in order of increasing energy (the electrons are getting farther from the nucleus). The order of levels filled looks like this:
1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, and 7p
One way to remember this pattern, probably the easiest, is to refer to the periodic table and remember where each orbital block falls to logically deduce this pattern. Another way is to make a table like the one below and use vertical lines to determine which subshells correspond with each other.
Pauli Exclusion Principle
The Pauli exclusion principle states that no two electrons can have the same four quantum numbers. The first three (n, l, and ml) may be the same, but the fourth quantum number must be different. A single orbital can hold a maximum of two electrons, which must have opposing spins; otherwise they would have the same four quantum numbers, which is forbidden. One electron is spin up (ms = +1/2) and the other would spin down (ms = -1/2). This tells us that each subshell has double the electrons per orbital. The s subshell has 1 orbital that can hold up to 2 electrons, the p subshell has 3 orbitals that can hold up to 6 electrons, the d subshell has 5 orbitals that hold up to 10 electrons, and the f subshell has 7 orbitals with 14 electrons.
Example 1: Hydrogen and Helium
The first three quantum numbers of an electron are n=1, l=0, ml=0. Only two electrons can correspond to these, which would be either ms = -1/2 or ms = +1/2. As we already know from our studies of quantum numbers and electron orbitals, we can conclude that these four quantum numbers refer to the 1s subshell. If only one of the ms values are given then we would have 1s1 (denoting hydrogen) if both are given we would have 1s2 (denoting helium). Visually, this is be represented as:
As shown, the 1s subshell can hold only two electrons and, when filled, the electrons have opposite spins.
Hund's Rule
When assigning electrons in orbitals, each electron will first fill all the orbitals with similar energy (also referred to as degenerate) before pairing with another electron in a half-filled orbital. Atoms at ground states tend to have as many unpaired electrons as possible. When visualizing this processes, think about how electrons are exhibiting the same behavior as the same poles on a magnet would if they came into contact; as the negatively charged electrons fill orbitals they first try to get as far as possible from each other before having to pair up.
Example 2: Oxygen and Nitrogen
If we look at the correct electron configuration of the Nitrogen (Z = 7) atom, a very important element in the biology of plants: 1s2 2s2 2p3
We can clearly see that p orbitals are half-filled as there are three electrons and three p orbitals. This is because Hund's Rule states that the three electrons in the 2p subshell will fill all the empty orbitals first before filling orbitals with electrons in them. If we look at the element after Nitrogen in the same period, Oxygen (Z = 8) its electron configuration is: 1s2 2s2 2p4 (for an atom).
Oxygen has one more electron than Nitrogen and as the orbitals are all half filled the electron must pair up.
The Aufbau Process
Aufbau comes from the German word "aufbauen" meaning "to build." When writing electron configurations, orbitals are built up from atom to atom. When writing the electron configuration for an atom, orbitals are filled in order of increasing atomic number. However, there are some exceptions to this rule.
Example 3: 3rd row elements
Following the pattern across a period from B (Z=5) to Ne (Z=10), the number of electrons increases and the subshells are filled. This example focuses on the p subshell, which fills from boron to neon.
• B (Z=5) configuration: 1s2 2s2 2p1
• C (Z=6) configuration:1s2 2s2 2p2
• N (Z=7) configuration:1s2 2s2 2p3
• O (Z=8) configuration:1s2 2s2 2p4
• F (Z=9) configuration:1s2 2s2 2p5
• Ne (Z=10) configuration:1s2 2s2 2p6
Example
The electron configuration for sulfur is 1s2 2s2 2p6 3s2 3p4 and can be represented using the orbital diagram below.
Exercises
Write the electron configuration for phosphorus and draw the orbital diagram.
Solution:
The electron configuration for phosphorus is 1s2 2s2 2p6 3s2 3p3 and the orbital diagram is drawn below. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/01%3A_Introduction_and_Review/1.04%3A_Electron_Configurations_and_Electronic_Orbital_Diagrams_%28Review%29.txt |
Learning Objective
Draw, interpret, and convert between Lewis (Kekule), Condensed, and Bond-line Structures
Note: The review of general chemistry in sections 1.3 - 1.6 is integrated into the above Learning Objective for organic chemistry in sections 1.7 and 1.8.
For organic chemistry, the emphasis is on the chemistry of carbon. The chemistry of carbon becomes more interesting when carbon is bonded to oxygen and/or nitrogen or other heteroatoms, atoms that are NOT carbon or hydrogen. Therefore, the octet rule is a strong factor in organic chemistry and is only violated by non-carbon elements like hydrogen, boron, aluminum, sulfur, and phosphorus.
Why are some substances chemically bonded molecules and others are an association of ions? The answer to this question depends upon the electronic structures of the atoms and nature of the chemical forces within the compounds. Although there are no sharply defined boundaries, chemical bonds are typically classified into three main types: ionic bonds, covalent bonds, and metallic bonds. In this chapter, each type of bond will be discussed and the general properties found in typical substances in which the bond type occurs
1. Ionic bonds results from electrostatic forces that exist between ions of opposite charge. These bonds typically involves a metal with a nonmetal
2. Covalent bonds result from the sharing of electrons between two atoms. The bonds typically involves one nonmetallic element with another
3. Metallic bonds These bonds are found in solid metals (copper, iron, aluminum) with each metal bonded to several neighboring groups and bonding electrons free to move throughout the 3-dimensional structure.
The Octet Rule
In 1904, Richard Abegg formulated what is now known as Abegg's rule, which states that the difference between the maximum positive and negative valences of an element is frequently eight. This rule was used later in 1916 when Gilbert N. Lewis formulated the "octet rule" in his cubical atom theory. The octet rule refers to the tendency of atoms to prefer to have eight electrons in the valence shell. When atoms have fewer than eight electrons, they tend to react and form more stable compounds. Atoms will react to get in the most stable state possible. A complete octet is very stable because all orbitals will be full. Atoms with greater stability have less energy, so a reaction that increases the stability of the atoms will release energy in the form of heat or light ;reactions that decrease stability must absorb energy, getting colder.
The Octet Rule: Atoms often gain, lose, or share electrons to achieve the same number of electrons as the noble gas closest to them in the periodic table.
When discussing the octet rule, we do not consider d or f electrons. Only the s and p electrons are involved in the octet rule, making it a useful rule for the main group elements (elements not in the transition metal or inner-transition metal blocks); an octet in these atoms corresponds to an electron configurations ending with s2p6.
Covalent Bonds
Covalent bonds form when atoms share electrons. Hydrogen is a first shell element with only one valence electron, so it can only form one bond creating a duet, an exception to the octet rule. With its four valence electrons, carbon can form four bonds to create an octet.
1. Normally two electrons pairs up and forms a bond, e.g., \(H_2\)
2. For most atoms there will be a maximum of eight electrons in the valence shell (octet structure), e.g., \(CH_4\)
The other tendency of atoms is to maintain a neutral charge. Only the noble gases (the elements on the right-most column of the periodic table) have zero charge with filled valence octets. All of the other elements have a charge when they have eight electrons all to themselves. The result of these two guiding principles is the explanation for much of the reactivity and bonding that is observed within atoms: atoms seek to share electrons in a way that minimizes charge while fulfilling an octet in the valence shell.
Ionic Bonds
Some atoms do not share electrons. Energetically, it is more favorable to fully gain or lose electrons to form ions. Ionic compounds form through the eletrostatic attraction of the ions to create a crystal lattice.
The formula for table salt is NaCl. It is the result of Na+ ions and Cl- ions bonding together. If sodium metal and chlorine gas mix under the right conditions, they will form salt. The sodium loses an electron, and the chlorine gains that electron. In the process, a great amount of light and heat is released. The resulting salt is mostly unreactive — it is stable. It will not undergo any explosive reactions, unlike the sodium and chlorine that it is made of. Why? Referring to the octet rule, atoms attempt to get a noble gas electron configuration, which is eight valence electrons. Sodium has one valence electron, so giving it up would result in the same electron configuration as neon. Chlorine has seven valence electrons, so if it takes one it will have eight (an octet). Chlorine has the electron configuration of argon when it gains an electron.
The octet rule could have been satisfied if chlorine gave up all seven of its valence electrons and sodium took them. In that case, both would have the electron configurations of noble gasses, with a full valence shell. However, their charges would be much higher. It would be Na7- and Cl7+, which is much less stable than Na+ and Cl-. Atoms are more stable when they have no charge, or a small charge.
Ionic Bonds Example
Lewis dot symbols can also be used to represent the ions in ionic compounds. The reaction of cesium with fluorine, for example, to produce the ionic compound CsF can be written as follows:
No dots are shown on Cs+ in the product because cesium has lost its single valence electron to fluorine. The transfer of this electron produces the Cs+ ion, which has the valence electron configuration of Xe, and the F ion, which has a total of eight valence electrons (an octet) and the Ne electron configuration. This description is consistent with the statement that among the main group elements, ions in simple binary ionic compounds generally have the electron configurations of the nearest noble gas. The charge of each ion is written in the product, and the anion and its electrons are enclosed in brackets. This notation emphasizes that the ions are associated electrostatically; no electrons are shared between the two elements.
Noble Gases
The noble gases rarely form compounds. They have the most stable configuration (full octet, no charge), so they have no reason to react and change their configuration. All other elements attempt to gain, lose, or share electrons to achieve a noble gas configuration.
Summary
Lewis dot symbols can be used to predict the number of bonds formed by most elements in their compounds. One convenient way to predict the number and basic arrangement of bonds in compounds is by using Lewis electron dot symbols, which consist of the chemical symbol for an element surrounded by dots that represent its valence electrons, grouped into pairs often placed above, below, and to the left and right of the symbol. The structures reflect the fact that the elements in period 2 and beyond tend to gain, lose, or share electrons to reach a total of eight valence electrons in their compounds, the so-called octet rule. Hydrogen, with only two valence electrons, does not obey the octet rule.
Lewis
Contributors and Attributions
• National Programme on Technology Enhanced Learning (India) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/01%3A_Introduction_and_Review/1.05%3A_Octet_Rule_-_Ionic_and_Covalent_Bonding_%28Review%29.txt |
Learning Objective
Draw, interpret, and convert between Lewis (Kekule), Condensed, and Bond-line Structures
Note: The review of general chemistry in sections 1.3 - 1.6 is integrated into the above Learning Objective for organic chemistry in sections 1.7 and 1.8.
Lewis Structures
Lewis structures, also known as Lewis-dot diagrams, show the bonding relationship between atoms of a molecule and the lone pairs of electrons in a molecule. While it can be helpful initially to write the individual shared electrons, this approach quickly becomes awkward.
A single line is used to represent one pair of shared electrons. Line representations are only used for shared electrons. Lone pair (unshared) electrons are still shown as individual electrons. Double and triple bonds can also be communicated with lines as shown below.
2 shared electrons form a single bond shown as ‘:’ or ‘–‘
4 shared electrons form a double bond shown as ‘::’ or ‘=’
6 shared electrons form at triple bond shown as ‘:::’ or
Unshared electrons are also called ‘Lone Pairs’ and are shown as ‘:’
Drawing Lone Pairs
Since the lone pair electrons are often NOT shown in chemical structures, it is important to mentally add the lone pairs. In the beginning, it can be helpful to physically add the lone pair electrons.
Bonding Patterns
For organic chemistry, the common bonding patterns of carbon, oxygen, and nitrogen have useful applications when evaluating chemical structures and reactivity.
Formal charges
Organic molecules can also have positive or negative charges associated with them. During chemical reactions, it is common to have charge reactant, intermediates, and/or products. Recognizing and distinguishing between neutral and charged bonding patterns will be helpful in learning reaction mechanisms. Consider the Lewis structure of methanol, CH3OH (methanol is the so-called ‘wood alcohol’ that unscrupulous bootleggers sometimes sold during the prohibition days in the 1920's, often causing the people who drank it to go blind). Methanol itself is a neutral molecule, but can lose a proton to become a molecular anion (CH3O-), or gain a proton to become a molecular cation (CH3OH2+).
The molecular anion and cation have overall charges of -1 and +1, respectively. But we can be more specific than that - we can also state for each molecular ion that a formal charge is located specifically on the oxygen atom, rather than on the carbon or any of the hydrogen atoms.
Figuring out the formal charge on different atoms of a molecule is a straightforward process - it’s simply a matter of adding up valence electrons.
A unbound oxygen atom has 6 valence electrons. When it is bound as part of a methanol molecule, however, an oxygen atom is surrounded by 8 valence electrons: 4 nonbonding electrons (two 'lone pairs') and 2 electrons in each of its two covalent bonds (one to carbon, one to hydrogen). In the formal charge convention, we say that the oxygen 'owns' all 4 nonbonding electrons. However, it only 'owns' one electron from each of the two covalent bonds, because covalent bonds involve the sharing of electrons between atoms. Therefore, the oxygen atom in methanol owns 2 + 2 + (½ x 4) = 6 valence electrons.
The formal charge on an atom is calculated as the number of valence electrons owned by the isolated atom minus the number of valence electrons owned by the bound atom in the molecule:
Determining formal charge on an atom
formal charge =
(number of valence electrons owned by the isolated atom)
- (number of valence electrons owned by the bound atom)
or . . .
formal charge =
(number of valence electrons owned by the isolated atom)
- (number of non-bonding electrons on the bound atom)
- ( ½ the number of bonding electrons on the bound atom)
Using this formula for the oxygen atom of methanol, we have:
formal charge on oxygen =
(6 valence electrons on isolated atom)
- (4 non-bonding electrons)
- (½ x 4 bonding electrons)
= 6 - 4 - 2 = 0. Thus, oxygen in methanol has a formal charge of zero (in other words, it has no formal charge).
How about the carbon atom in methanol? An isolated carbon owns 4 valence electrons. The bound carbon in methanol owns (½ x 8) = 4 valence electrons:
formal charge on carbon =
(4 valence electron on isolated atom)
- (0 nonbonding electrons)
- (½ x 8 bonding electrons)
= 4 - 0 - 4 = 0. So the formal charge on carbon is zero.
For each of the hydrogens in methanol, we also get a formal charge of zero:
formal charge on hydrogen =
(1 valence electron on isolated atom)
- (0 nonbonding electrons)
- (½ x 2 bonding electrons)
= 1 - 0 - 1 = 0
Now, let's look at the cationic form of methanol, CH3OH2+. The bonding picture has not changed for carbon or for any of the hydrogen atoms, so we will focus on the oxygen atom.
The oxygen owns 2 non-bonding electrons and 3 bonding elections, so the formal charge calculations becomes:
formal charge on oxygen =
(6 valence electrons in isolated atom)
- (2 non-bonding electrons)
- (½ x 6 bonding electrons)
= 6 - 2 - 3 = 1. A formal charge of +1 is located on the oxygen atom.
For methoxide, the anionic form of methanol, the calculation for the oxygen atom is:
formal charge on oxygen =
(6 valence electrons in isolated atom)
- (6 non-bonding electrons)
- (½ x 2 bonding electrons)
= 6 - 6 - 1 = -1. A formal charge of -1 is located on the oxygen atom.
A very important rule to keep in mind is that the sum of the formal charges on all atoms of a molecule must equal the net charge on the whole molecule.
When drawing the structures of organic molecules, it is very important to show all non-zero formal charges, being clear about where the charges are located. A structure that is missing non-zero formal charges is not correctly drawn, and will probably be marked as such on an exam!
At this point, thinking back to what you learned in general chemistry, you are probably asking “What about dipoles? Doesn’t an oxygen atom in an O-H bond ‘own’ more of the electron density than the hydrogen, because of its greater electronegativity?” This is absolutely correct, and we will be reviewing the concept of bond dipoles later on. For the purpose of calculating formal charges, however, bond dipoles don’t matter - we always consider the two electrons in a bond to be shared equally, even if that is not an accurate reflection of chemical reality. Formal charges are just that - a formality, a method of electron book-keeping that is tied into the Lewis system for drawing the structures of organic compounds and ions. Later, we will see how the concept of formal charge can help us to visualize how organic molecules react.
Finally, don't be lured into thinking that just because the net charge on a structure is zero there are no atoms with formal charges: one atom could have a positive formal charge and another a negative formal charge, and the net charge would still be zero. Zwitterions, such as amino acids, have both positive and negative formal charges on different atoms:
Even though the net charge on glycine is zero, it is still neccessary to show the location of the positive and negative formal charges.
Exercise 1.4
Fill in all missing lone pair electrons and formal charges in the structures below. Assume that all atoms have a complete valence shell of electrons. Net charges are shown outside the brackets.
Solutions to exercises | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/01%3A_Introduction_and_Review/1.06%3A_Lewis_Structures_and_Formal_Charges_%28Review%29.txt |
Learning Objective
Draw, interpret, and convert between Lewis (Kekule), Condensed, and Bond-line Structures
Note: The review of general chemistry in sections 1.3 - 1.6 is integrated into the above Learning Objective for organic chemistry in sections 1.7 and 1.8.
Common bonding patterns in organic structures
The Lewis structure below (of one of the four nucleoside building blocks that make up DNA) can appear complex and confusing at first glance. Fortunately, common bonding patterns occur that can allow for simplifications when drawing structures. The rules for the simplifies structures rely on the neutral bonding patterns for carbon, orygen, nitrogen, phosphorus, and sulfur primarily. Since organic compounds have a hydrocarbon backbone, the atoms that are NOT carbon and hydrogen are called heteroatoms. 2'-deoxycytidine contains seven heteroatoms: four oxygen atoms land three nitrogen atoms. Heteroatoms are a primary source of chemical reactivity for organic chemistry.
Heteroatoms: atoms in an organic compound that are NOT carbon or hydrogen,
typically oxygen, nitrogen, phosphorus, and sulfur
The ability to quickly and efficiently draw large structures and determine formal charges is not terribly hard to come by - all it takes is a few shortcuts and some practice at recognizing common bonding patterns.
Let’s start with carbon, the most important element for organic chemists. Carbon is said to be tetravalent, meaning that it tends to form four bonds. If you look at the simple structures of methane, methanol, ethane, ethene, and ethyne in the figures from the previous section, you should quickly recognize that in each molecule, the carbon atom has four bonds, and a formal charge of zero.
This is a pattern that holds throughout most of the organic molecules we will see, but there are also exceptions.
<style type="text/css"></style> In carbon dioxide, the carbon atom has double bonds to oxygen on both sides (O=C=O). Later on in this chapter and throughout this book we will see examples of organic ions called ‘carbocations’ and carbanions’, in which a carbon atom bears a positive or negative formal charge, respectively. If a carbon has only three bonds and an unfilled valence shell (in other words, if it does not fulfill the octet rule), it will have a positive formal charge.
If, on the other hand, it has three bonds plus a lone pair of electrons, it will have a formal charge of -1. Another possibility is a carbon with three bonds and a single, unpaired (free radical) electron: in this case, the carbon has a formal charge of zero. (One last possibility is a highly reactive species called a ‘carbene’, in which a carbon has two bonds and one lone pair of electrons, giving it a formal charge of zero. You may encounter carbenes in more advanced chemistry courses, but they will not be discussed any further in this book).
You should certainly use the methods you have learned to check that these formal charges are correct for the examples given above. More importantly, you will need, before you progress much further in your study of organic chemistry, to simply recognize these patterns (and the patterns described below for other atoms) and be able to identify carbons that bear positive and negative formal charges by a quick inspection.
The pattern for hydrogens is easy: hydrogen atoms have only one bond, and no formal charge. The exceptions to this rule are the proton, H+, and the hydride ion, H-, which is a proton plus two electrons. Because we are concentrating in this book on organic chemistry as applied to living things, however, we will not be seeing ‘naked’ protons and hydrides as such, because they are too reactive to be present in that form in aqueous solution. Nonetheless, the idea of a proton will be very important when we discuss acid-base chemistry, and the idea of a hydride ion will become very important much later in the book when we discuss organic oxidation and reduction reactions. As a rule, though, all hydrogen atoms in organic molecules have one bond, and no formal charge.
Let us next turn to oxygen atoms. Typically, you will see an oxygen bonding in three ways, all of which fulfill the octet rule.
If it has two bonds and two lone pairs, as in water, it will have a formal charge of zero. If it has one bond and three lone pairs, as in hydroxide ion, it will have a formal charge of-1. If it has three bonds and one lone pair, as in hydronium ion, it will have a formal charge of +1.
When we get to our discussion of free radical chemistry in chapter 17, we will see other possibilities, such as where an oxygen atom has one bond, one lone pair, and one unpaired (free radical) electron, giving it a formal charge of zero. For now, however, concentrate on the three main non-radical examples, as these will account for virtually everything we see until chapter 17.
Nitrogen has two major bonding patterns, both of which fulfill the octet rule:
If a nitrogen has three bonds and a lone pair, it has a formal charge of zero. If it has four bonds (and no lone pair), it has a formal charge of +1. In a fairly uncommon bonding pattern, negatively charged nitrogen has two bonds and two lone pairs.
Two third row elements are commonly found in biological organic molecules: sulfur and phosphorus. Although both of these elements have other bonding patterns that are relevant in laboratory chemistry, in a biological context sulfur almost always follows the same bonding/formal charge pattern as oxygen, while phosphorus is present in the form of phosphate ion (PO43-), where it has five bonds (almost always to oxygen), no lone pairs, and a formal charge of zero. Remember that atoms of elements in the third row and below in the periodic table have 'expanded valence shells' with d orbitals available for bonding, and the the octet rule does not apply.
Finally, the halogens (fluorine, chlorine, bromine, and iodine) are very important in laboratory and medicinal organic chemistry, but less common in naturally occurring organic molecules. Halogens in organic compounds usually are seen with one bond, three lone pairs, and a formal charge of zero. Sometimes, especially in the case of bromine, we will encounter reactive species in which the halogen has two bonds (usually in a three-membered ring), two lone pairs, and a formal charge of +1.
These rules, if learned and internalized so that you don’t even need to think about them, will allow you to draw large organic structures, complete with formal charges, quite quickly.
Once you have gotten the hang of drawing Lewis structures, it is not always necessary to draw lone pairs on heteroatoms, as you can assume that the proper number of electrons are present around each atom to match the indicated formal charge (or lack thereof). Occasionally, though, lone pairs are drawn if doing so helps to make an explanation more clear.
Exercise 1: Draw one structure that corresponds to each of the following molecular formulas, using the common bonding patters covered above. Be sure to include all lone pairs and formal charges where applicable, and assume that all atoms have a full valence shell of electrons. More than one correct answer is possible for each, so you will want to check your answers with your instructor or tutor.
a) C5H10O b) C5H8O c) C6H8NO+ d) C4H3O2-
Solutions to exercises | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/01%3A_Introduction_and_Review/1.07%3A_Common_Bonding_Patterns_for_Organic_Chemistry.txt |
Learning Objective
Draw, interpret, and convert between Lewis (Kekule), Condensed, and Bond-line Structures
Note: The review of general chemistry in sections 1.3 - 1.6 is integrated into the above Learning Objective for organic chemistry in sections 1.7 and 1.8.
Shorthand notations to represent organic molecules rely on our knowledge of common neutral bonding patterns. Knowing these patterns, we can fill in the missing structural information. Some of these shorthand ways of drawing molecules give us insight into the bond angles and relative positions of atoms in the molecule, while some notations eliminate the carbon and hydrogen atoms and only indicate the heteroatoms (the atoms that are NOT carbon or hydrogen).
There are three primary methods to communicate chemical structure of organic molecules:
Kekule: Lewis structures using lines to represent covalent bonds and showing all atoms and lone pair electrons
Bond-line (Skeletl-line): shows bonds between carbon atoms and heteroatoms) (with lone pair electrons when requested)
Condensed: all atoms are written to communicate structure without drawing any chemical bonds based on the carbon backbone
Introduction
Observe the following drawings of the structure of Retinol, the most common form of vitamin A. The first drawing follows the straight-line (a.k.a. Kekulé) structure which is helpful when you want to look at every single atom; however, showing all of the hydrogen atoms makes it difficult to compare the overall structure with other similar molecules and makes it difficult to focus in on the double bonds and OH group.
Retinol: Kekulé straight-line drawing
The following is a bond-line (a.k.a. zig-zag) formula for retinol. With this simiplified representation, one can easily see the carbon-carbon bonds, double bonds, OH group, and CH3 groups sticking off of the the main ring and chain. Also, it is much quicker to draw this than the one above. You will learn to appreciate this type of formula writing after drawing a countless number of organic molecules.
Retinol: Bond-line or zig-zag formula
Importance of Structure
Learning and practicing the basics of Organic Chemistry will help you immensely in the long run as you learn new concepts and reactions. Some people say that Organic Chemistry is like another language, and in some aspects, it is. At first it may seem difficult or overwhelming, but the more you practice looking at and drawing organic molecules, the more familiar you will become with the structures and formulas. Another good idea is to get a model kit and physically make the molecules that you have trouble picturing in your head.
Through general chemistry, you may have already experienced looking at molecular structure. The different ways to draw organic molecules include Keku (straight-line), Condensed Formulas, and Bond-Line Formulas (zig-zag). It will be more helpful if you become comfortable going from one style of drawing to another, and look at drawings and understanding what they mean, than knowing which kind of drawing is named what.
An example of a drawing that incorporates all three ways to draw organic molecules would be the following additional drawing of Retinol. The majority of the drawing is Bond-line (zig-zag) formula, but the -CH3 are written as condensed formulas, and the -OH group is written in Kekulé form.
A widely used way of showing the 3D structure of molecules is the use of dashes, wedges, and straight lines. This drawing method is essential because the placement of different atoms could yield different molecules even if the molecular formulas were exactly the same. Below are two drawings of a 4-carbon molecule with two chlorines and two bromines attached.
4-carbon molecule with 2 chlorines and 2 bromines 4-carbon molecule with 2 chlorines and 2 bromines
Both drawings look like they represent the same molecule; however, if we add dashes and wedged we will see that two different molecules could be depicted:
The two molecules above are different, prove this to yourself by building a model. An easier way to compare the two molecules is to rotate one of the bonds (here, it is the bond on the right):
Notice how the molecule on the right has both bromines on the same side and chlorines on the same side, whereas the first molecule is different. Read about Dashed-Wedged Line structures, bottom of page, to understand what has been introduced above. You will learn more about the importance of atomic connectivity in molecules as you continue on to learn about Stereochemistry.
Drawing the Structure of Organic Molecules
Although larger molecules may look complicated, they can be easily understood by breaking them down and looking at their smaller components.
All atoms want to have their valence shell full, a "closed shell." Hydrogen wants to have 2 e- whereas carbon, oxygen, and nitrogen want to have 8 e-. When looking at the different representations of molecules, keep in mind the Octet Rule. Also remember that hydrogen can bond one time, oxygen can bond up to two times, nitrogen can bond up to three times, and carbon can bond up to four times.
Kekulé (a.k.a. Lewis Structures)
Kekulé structures are similar to Lewis Structures, but instead of covalent bonds being represented by electron dots, the two shared electrons are shown by a line.
(A) (B)(C)
Lone pairs remain as two electron dots, or are sometimes left out even though they are still there. Notice how the three lone pairs of electrons were not draw in around chlorine in example B.
Condensed Formulas
A condensed formula is made up of the elemental symbols. The order of the atoms suggests the connectivity. Condensed formulas can be read from either direction and H3C is the same as CH3, although the latter is more common because Look at the examples below and match them with their identical molecule under Kekulé structures and bond-line formulas.
(A) CH3CH2OH (B) ClCH2CH2CH(OCH3)CH3 (C) H3CNHCH2COOH
Let's look closely at example B. As you go through a condensed formula, you want to focus on the carbons and other elements that aren't hydrogen. The hydrogen's are important, but are usually there to complete octets. Also, notice the -OCH3 is in written in parentheses which tell you that it not part of the main chain of carbons. As you read through a a condensed formula, if you reach an atom that doesn't have a complete octet by the time you reach the next hydrogen, then it's possible that there are double or triple bonds. In example C, the carbon is double bonded to oxygen and single bonded to another oxygen. Notice how COOH means C(=O)-O-H instead of CH3-C-O-O-H because carbon does not have a complete octet and oxygens.
Bond-Line (a.k.a. zig-zag) Formulas
The name gives away how this formula works. This formula is full of bonds and lines, and because of the typical (more stable) bonds that atoms tend to make in molecules, they often end up looking like zig-zag lines. If you work with a molecular model kit you will find it difficult to make stick straight molecules (unless they contain sp triple bonds) whereas zig-zag molecules and bonds are much more feasible.
(A) (B) (C)
These molecules correspond to the exact same molecules depicted for Kekulé structures and condensed formulas. Notice how the carbons are no longer drawn in and are replaced by the ends and bends of a lines. In addition, the hydrogens have been omitted, but could be easily drawn in (see practice problems). Although we do not usually draw in the H's that are bonded to carbon, we do draw them in if they are connected to other atoms besides carbon (example is the OH group above in example A) . This is done because it is not always clear if the non-carbon atom is surrounded by lone pairs or hydrogens. Also in example A, notice how the OH is drawn with a bond to the second carbon, but it does not mean that there is a third carbon at the end of that bond/ line.
Dashed-Wedged Line Structure
As you may have guessed, the Dashed-Wedged Line structure is all about lines, dashes, and wedges. At first it may seem confusing, but with practice, understanding dash-wedged line structures will become like second nature. The following are examples of each, and how they can be used together.
Above are 4-carbon chains with attached OH groups or Cl and Br atoms. Remember that each line represents a bond and that the carbons and hydrogens have been omitted. When you look at or draw these structures, the straight lines illustrate atoms and bonds that are in the same plane, the plane of the paper (in this case, computer screen). Dashed lines show atoms and bonds that go into the page, behind the plane, away from you. In the above example, the OH group is going into the plane, while at the same time a hydrogen comes out (wedged).
Blue bead= OH group; White bead=H
Wedged lines illustrate bonds and atoms that come out of the page, in front of the plane, toward you. In the 2D diagram above, the OH group is coming out of the plane of the paper, while a hydrogen goes in (dashed).
Blue bead= OH group; White bead=H
As stated before, straight lines illustrate atoms and bonds that are in the same plane as the paper, but in the 2D example, the straight line bond for OH means that it it unsure or irrelevant whether OH is going away or toward you. It is also assumed that hydrogen is also connected to the same carbon that OH is on.
Blue bead= OH group; H is not shown
Try using your model kit to see that the OH group cannot lie in the same plane at the carbon chain (don't forget your hydrogens!). In the final 2Dexample, both dashed and wedged lines are used because the attached atoms are not hydrogens (although dashed and wedged lines can be used for hydrogens).The chlorine is coming out the page while bromine is going into the page.
Blue bead=Cl; Red bead=Br
Example: Converting between Structural Formulas
Throughout the course, it will be helpful to convert compounds into different structural formulas (Kekule (Lewis Structures), Bond-line, and Condensed) depending on the type of question that is asked. Standardized exams frequently include a high percentage of condensed formulas because it is easier and cheaper to type letters and numbers than to import figures. Initially, it can be tricky writing a bond-line structure directly from a condensed formula. First write the Kekule structure from the condensed formula and then draw the bond-line structure from the Kekule. Practice will quickly allow you to convert directly between condensed and bond-line structures.
The condensed formula for propanal is CH3CH2CHO. Can you visualize the bond-line structure of propanal? If yes, excellent, If no, the following practice will help.
The Kekule structure for propanal is shown below.
The bond-line structure for propanal is shown below.
All three structures represent the same compound, propanal.
Exercises
1. How many carbons are in the following drawing? How many hydrogens?
2. How many carbons are in the following drawing? How many hydrogens?
3. How many carbons are in the following drawing? How many hydrogens?
4. Look at the following molecule of vitamin A and draw in the hidden hydrogens and electron pairs.
(hint: Do all of the carbons have 4 bonds? Do all the oxygens have a full octet?)
5. How many bonds can hydrogen make?
6. How many bonds can chlorine make?
7. Dashed lines means the atomic bond goes ___________(away/toward) you.
8. Draw ClCH2CH2CH(OCH3)CH3 in Kekuléand zig-zag form.
9. Extra practice problems can be found ______?
Solutions
1. Remember the octet rule and how many times carbons and hydrogens are able to bond to other atoms.
2. Electron pairs drawn in blue and hydrogens draw in red.
3. Hygrogen can make one bond.
4. Chlorine can make one bond.
5. Away
6. See (B) under Kekulé and Bond-line (zig-zag) formulas.
7. Extra practice problems can be found: in your textbook, homework, lecture notes, online, reference books, and more. Try making up some of your own molecules, they may exist!
Contributors and Attributions
• Choo, Ezen (2009, UCD '11)
The building block of structural organic chemistry is the tetravalent carbon atom. With few exceptions, carbon compounds can be formulated with four covalent bonds to each carbon, regardless of whether the combination is with carbon or some other element. The two-electron bond, which is illustrated by the carbon-hydrogen bonds in methane or ethane and the carbon-carbon bond in ethane, is called a single bond. In these and many related substances, each carbon is attached to four other atoms:
There exist, however, compounds such as ethene (ethylene), $C_2H_4$, in which two electrons from each of the carbon atoms are mutually shared, thereby producing two two-electron bonds, an arrangement which is called a double bond. Each carbon in ethene is attached to only three other atoms:
Similarly, in ethyne (acetylene), $C_2H_2$, three electrons from each carbon atom are mutually shared, producing three two-electron bonds, called a triple bond, in which each carbon is attached to only two other atoms:
Of course, in all cases each carbon has a full octet of electrons. Carbon also forms double and triple bonds with several other elements that can exhibit a covalence of two or three. The carbon-oxygen (or carbonyl) double bond appears in carbon dioxide and many important organic compounds such as methanal (formaldehyde) and ethanoic acid (acetic acid). Similarly, a carbon-nitrogen triple bond appears in methanenitrile (hydrogen cyanide) and ethanenitrile (acetonitrile).
By convention, a single straight line connecting the atomic symbols is used to represent a single (two-electron) bond, two such lines to represent a double (four-electron) bond, and three lines a triple (six-electron) bond. Representations of compounds by these symbols are called structural formulas; some examples are
A point worth noting is that structural formulas usually do not indicate the nonbonding electron pairs. This is perhaps unfortunate because they play as much a part in the chemistry of organic molecules as do the bonding electrons and their omission may lead the unwary reader to overlook them. However, when it is important to represent them, this can be done best with pairs of dots, although a few authors use lines:
To save space and time in the representation of organic structures, it is common practice to use "condensed formulas" in which the bonds are not shown explicitly. In using condensed formulas, normal atomic valences are understood throughout. Examples of condensed formulas are
Another type of abbreviation that often is used, particularly for ring compounds, dispenses with the symbols for carbon and hydrogen atoms and leaves only the lines in a structural formula. For instance, cyclopentane, $C_5H_{10}$, often is represented as a regular pentagon in which it is understood that each apex represents a carbon atom with the requisite number of hydrogens to satisfy the tetravalence of carbon:
Likewise, cyclopropane, $C_3H_6$; cyclobutane, $C_4H_8$; and cyclohexane, $C_6H_{12}$, are drawn as regular polygons:
Although this type of line drawing is employed most commonly for cyclic structures, its use for open chain (acyclic) structures is becoming increasingly widespread. There is no special merit to this abbreviation for simple structures such as butane, $C_4H_{10}$; 1-butene, $C_4H_8$; or 1,3-butadiene, $C_4H_6$, but it is of value in representing more complex molecules such as $\beta$-carotene, $C_{40}H_{56}$:
Line structures also can be modified to represent the three-dimensional shapes of molecules, and the way that this is done will be discussed in detail in Chapter 5. At the onset of you study of organic chemistry, you should write out the formulas rather completely until you are thoroughly familiar with what these abbreviations stand for.
Contributors and Attributions
John D. Robert and Marjorie C. Caserio (1977) Basic Principles of Organic Chemistry, second edition. W. A. Benjamin, Inc. , Menlo Park, CA. ISBN 0-8053-8329-8. This content is copyrighted under the following conditions, "You are granted permission for individual, educational, research and non-commercial reproduction, distribution, display and performance of this work in any format." | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/01%3A_Introduction_and_Review/1.08%3A_Structural_Formulas_-_Lewis_Kekule_Bond-line_Condensed_and_Perspective.txt |
Learning Objective
• Identify polar bonds and compounds
Electronegativity is a measure of the tendency of an atom to attract a bonding pair of electrons. The Pauling scale is the most commonly used. Fluorine (the most electronegative element) is assigned a value of 4.0, and values range down to cesium and francium which are the least electronegative at 0.7.
Patterns of electronegativity in the Periodic Table
Electronegativity is defined as the ability of an atom in a particular molecule to attract electrons to itself. The greater the value, the greater the attractiveness for electrons.
Trends in electronegativity across a period
The positively charged protons in the nucleus attract the negatively charged electrons. As the number of protons in the nucleus increases, the electronegativity or attraction will increase. Therefore electronegativity increases from left to right in a row in the periodic table. This effect only holds true for a row in the periodic table because the attraction between charges falls off rapidly with distance. The chart shows electronegativities from sodium to chlorine (ignoring argon since it does not does not form bonds).
Trends in electronegativity down a group
As you go down a group, electronegativity decreases. (If it increases up to fluorine, it must decrease as you go down.) The chart shows the patterns of electronegativity in Groups 1 and 7.
Explaining the patterns in electronegativity
The attraction that a bonding pair of electrons feels for a particular nucleus depends on:
• the number of protons in the nucleus;
• the distance from the nucleus;
• the amount of screening by inner electrons.
Why does electronegativity increase across a period?
Consider sodium at the beginning of period 3 and chlorine at the end (ignoring the noble gas, argon). Think of sodium chloride as if it were covalently bonded.
Both sodium and chlorine have their bonding electrons in the 3-level. The electron pair is screened from both nuclei by the 1s, 2s and 2p electrons, but the chlorine nucleus has 6 more protons in it. It is no wonder the electron pair gets dragged so far towards the chlorine that ions are formed. Electronegativity increases across a period because the number of charges on the nucleus increases. That attracts the bonding pair of electrons more strongly.
Why does electronegativity fall as you go down a group?
As you go down a group, electronegativity decreases because the bonding pair of electrons is increasingly distant from the attraction of the nucleus. Consider the hydrogen fluoride and hydrogen chloride molecules:
The bonding pair is shielded from the fluorine's nucleus only by the 1s2 electrons. In the chlorine case it is shielded by all the 1s22s22p6 electrons. In each case there is a net pull from the center of the fluorine or chlorine of +7. But fluorine has the bonding pair in the 2-level rather than the 3-level as it is in chlorine. If it is closer to the nucleus, the attraction is greater.
Dipole moments occur when there is a separation of charge. They can occur between two ions in an ionic bond or between atoms in a covalent bond; dipole moments arise from differences in electronegativity. The larger the difference in electronegativity, the larger the dipole moment. The distance between the charge separation is also a deciding factor into the size of the dipole moment. The dipole moment is a measure of the polarity of the molecule.
Bond Polarity & Dipole Moment
Atoms with differences in electronegativity will share electrons unequally. The shared electrons of the covalent bond are held more tightly at the more electronegative element creating a partial negative charge, while the less electronegative element has a partial positive charge, . The larger the difference in electronegativity between the two atoms, the more polar the bond. To be considered a polar bond, the difference in electronegativity must >0.4 on the Pauling scale. Since the two electrical partial charges have opposite sign and equal magnitude and are separated by a distance, a dipole is established. Dipole moment is measured in debye units, which is equal to the distance between the charges multiplied by the charge (1 debye equals 3.34 x 10-30 coulomb-meters).
Polarity and Structure of Molecules
The shape of a molecule AND the polarity of its bonds. A molecule that contains polar bonds, might not have any overall polarity, depending upon its shape. The simple definition of whether a complex molecule is polar or not depends upon whether its overall centers of positive and negative charges overlap. If these centers lie at the same point in space, then the molecule has no overall polarity (and is non polar).
If a molecule is completely symmetric, then the dipole moment vectors on each molecule will cancel each other out, making the molecule nonpolar. A molecule can only be polar if the structure of that molecule is not symmetric.
A good example of a nonpolar molecule that contains polar bonds is carbon dioxide. This is a linear molecule and the C=O bonds are, in fact, polar. The central carbon will have a net positive charge, and the two outer oxygens a net negative charge. However, since the molecule is linear, these two bond dipoles cancel each other out (i.e. vector addition of the dipoles equals zero). And the overall molecule has no dipole (μ=0.
Although a polar bond is a prerequisite for a molecule to have a dipole, not all molecules with polar bonds exhibit dipoles
Geometric Considerations
Example 1: Polar Bonds vs. Polar Molecules
In a simple diatomic molecule like HCl, if the bond is polar, then the whole molecule is polar. What about more complicated molecules?
Consider CCl4, (left panel in figure above), which as a molecule is not polar - in the sense that it doesn't have an end (or a side) which is slightly negative and one which is slightly positive. The whole of the outside of the molecule is somewhat negative, but there is no overall separation of charge from top to bottom, or from left to right.
In contrast, CHCl3 is a polar molecule (right panel in figure above). The hydrogen at the top of the molecule is less electronegative than carbon and so is slightly positive. This means that the molecule now has a slightly positive "top" and a slightly negative "bottom", and so is overall a polar molecule.
A polar molecule will need to be "lop-sided" in some way.
Example 2: C2Cl4
Although the C–Cl bonds are rather polar, the individual bond dipoles cancel one another in this symmetrical structure, and Cl2C=CCl2 does not have a net dipole moment.
Example 3: CH3Cl
C-Cl, the key polar bond, is 178 pm. Measurement reveals 1.87 D. From this data, % ionic character can be computed. If this bond were 100% ionic (based on proton & electron),
μ=178100(4.80D)=8.54D
Example 4: HCl
Since measurement 1.87 D,
% ionic = (1.7/8.54)x100 = 22%HCl
u = 1.03 D (measured) H-Cl bond length 127 pm
If 100% ionic,
μ=127100(4.80D)=6.09D
ionic = (1.03/6.09)x100 = 17%
A "spectrum" of bonds
The implication of all this is that there is no clear-cut division between covalent and ionic bonds. In a pure covalent bond, the electrons are held on average exactly half way between the atoms. In a polar bond, the electrons have been dragged slightly towards one end. How far does this dragging have to go before the bond counts as ionic? There is no real answer to that. Sodium chloride is typically considered an ionic solid, but even here the sodium has not completely lost control of its electron. Because of the properties of sodium chloride, however, we tend to count it as if it were purely ionic. Lithium iodide, on the other hand, would be described as being "ionic with some covalent character". In this case, the pair of electrons has not moved entirely over to the iodine end of the bond. Lithium iodide, for example, dissolves in organic solvents like ethanol - not something which ionic substances normally do.
Summary
• No electronegativity difference between two atoms leads to a pure non-polar covalent bond.
• A small electronegativity difference leads to a polar covalent bond.
• A large electronegativity difference leads to an ionic bond.
Example 1: Polar Bonds vs. Polar Molecules
In a simple diatomic molecule like HCl, if the bond is polar, then the whole molecule is polar. What about more complicated molecules?
Consider CCl4, (left panel in figure above), which as a molecule is not polar - in the sense that it doesn't have an end (or a side) which is slightly negative and one which is slightly positive. The whole of the outside of the molecule is somewhat negative, but there is no overall separation of charge from top to bottom, or from left to right.
In contrast, CHCl3 is a polar molecule (right panel in figure above). The hydrogen at the top of the molecule is less electronegative than carbon and so is slightly positive. This means that the molecule now has a slightly positive "top" and a slightly negative "bottom", and so is overall a polar molecule.
A polar molecule will need to be "lop-sided" in some way.
Exercises
For the following compounds,
a) add lone pairs of electrons to complete octets
b) add dipole moment arrows or partial +/- signs to indicate polar bonds
c) predict the molecular polarity (Remember to visualize each compound in three dimensions.)
Solutions
Contributors and Attributions
• Jim Clark (Chemguide.co.uk)
• Prof. Richard Bank, Boise State University, Emeritus, | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/01%3A_Introduction_and_Review/1.09%3A_Electronegativity_and_Bond_Polarity_%28Review%29.txt |
Learning Objective
• Draw resonance forms and predict the relative contribution of each resonance form to the overall structure of the compound or ion
Recognizing resonance
Resonance contributors involve the ‘imaginary movement’ of pi-bonded electrons or of lone-pair electrons that are adjacent to (i.e. conjugated to) pi bonds. You can never shift the location of electrons in sigma bonds – if you show a sigma bond forming or breaking, you are showing a chemical reaction taking place. Likewise, the positions of atoms in the molecule cannot change between two resonance contributors.
Because benzene will appear throughout this course, it is important to recognize the stability gained through the resonance delocalization of the six pi electrons throughout the six carbon atoms. Benzene also illustrates one way to recognize resonance - when it is possible to draw two or more equivalent Lewis strucutres. If we were to draw the structure of an aromatic molecule such as 1,2-dimethylbenzene, there are two ways that we could draw the double bonds:
Which way is correct? There are two simple answers to this question: 'both' and 'neither one'. Both ways of drawing the molecule are equally acceptable approximations of the bonding picture for the molecule, but neither one, by itself, is an accurate picture of the delocalized pi bonds. The two alternative drawings, however, when considered together, give a much more accurate picture than either one on its own. This is because they imply, together, that the carbon-carbon bonds are not double bonds, not single bonds, but about halfway in between.
When it is possible to draw more than one valid structure for a compound or ion, we have identified resonance contributors: two or more different Lewis structures depicting the same molecule or ion that, when considered together, do a better job of approximating delocalized pi-bonding than any single structure. By convention, resonance contributors are linked by a double-headed arrow, and are sometimes enclosed by brackets:
In order to make it easier to visualize the difference between two resonance contributors, small, curved arrows are often used. Each of these arrows depicts the ‘movement’ of two pi electrons. A few chapters from now when we begin to study organic reactions - a process in which electron density shifts and covalent bonds between atoms break and form - this ‘curved arrow notation’ will become extremely important in depicting electron movement. In the drawing of resonance contributors, however, this electron ‘movement’ occurs only in our minds, as we try to visualize delocalized pi bonds. Nevertheless, use of the curved arrow notation is an essential skill that you will need to develop in drawing resonance contributors.
The depiction of benzene using the two resonance contributors A and B in the figure above does not imply that the molecule at one moment looks like structure A, then at the next moment shifts to look like structure B. Rather, at all moments, the molecule is a combination, or resonance hybrid of both A and B.
Caution! It is very important to be clear that in drawing two (or more) resonance contributors, we are not drawing two different molecules: they are simply different depictions of the exact same molecule. Furthermore, the double-headed resonance arrow does NOT mean that a chemical reaction has taken place.
Usually, derivatives of benzene (and phenyl groups, when the benzene ring is incorporated into a larger organic structure) are depicted with only one resonance contributor, and it is assumed that the reader understands that resonance hybridization is implied. This is the convention that will be used for the most part in this book. In other books or articles, you may sometimes see benzene or a phenyl group drawn with a circle inside the hexagon, either solid or dashed, as a way of drawing a resonance hybrid.
Curved arrows communicate electron flow (movement)
Curved arrows indicate electron flow. The base of the curved arrow is placed at the source of the electrons that are moving. The head of the arrow is placed at the destination of the electrons. A single barbed arrow represents one electron and a double barb represents two electrons. Electrons move from regions of relative high density to regions of low density or toward electronegative elements. It is important to use accuracy and precision when drawing curved arrows.
It is also important to consciously use the correct type of arrow. There are four primary types of arrows used by chemists to communicate one of the following: completion reaction, equilibrium reaction, electron movement, resonance forms. The three other types of arrows are shown below to build discernment between them.
Resonance Delocalizes Charge to Increase Stability
Resonance is most useful wen it delocalizes charge to stabilize reactive intermediates and products. Recognizing, drawing, and evaluating the relative stability of resonance contributors is essential to understanding organic reaction mechanisms.
Guidelines for drawing and working with resonance contributors
Learning to draw and interpret resonance structures, there are a few basic guidelines to help avoid drawing nonsensical structures. All of these guidellines make perfect sense as long as we remember that resonance contributors are merely a human-invented convention for depicting the delocalization of pi electrons in conjugated systems. When we see two different resonance contributors, we are not seeing a chemical reaction! We are seeing the exact same molecule or ion depicted in two different ways. All resonance contributors must be drawn as proper Lewis structures, with correct formal charges. Never show curved 'electron movement' arrows that would lead to a situation where a second-row element (ie. carbon, nitrogen, or oxygen) has more than eight electrons: this would break the 'octet rule'. Sometimes, however, we will draw resonance contributors in which a carbon atom has only six electrons (ie. a carbocation). In general, all oxygen and nitrogen atoms should have a complete octet of valence electrons.
1. There is ONLY ONE STRUCTURE for each compound or ion. This structure takes its character from the sum of all the contributors, not all resonance structures contribute equally to the sum.
2. Atoms must maintain their same position.
3. Only e- move !
4. All resonance contributors for a molecule or ion must have the same net charge.
5. Recognize which electrons can participate in resonance
a) unshared e- pairs or radicals
b) pi bond electrons
6. Recognize electron receptors
a) atoms with a positive (+) charge
b) electronegative atoms that can tolerate a negative charge
c) atoms which possess delocalizable electrons - see #4 above
7. Common electron flow patterns
a) move pi e- toward positive (+) charge or other pi bonds
b) move non-bonding e- pairs toward pi bonds
c) move single non-bonding e- toward pi bonds
Evaluating Resonance Contributors
1. Identical structures are equally important.
2. Structures will a greater number of bonds are more important.
3. Structures with charge separation are less important.
4. Pay attention to electronegativities.
5. Neutral atoms need to have complete octets.
Resonance Contributors for the Carboxylate Group
The convention of drawing two or more resonance contributors to approximate a single structure may seem a bit clumsy to you at this point, but as you gain experience you will see that the practice is actually very useful when discussing the manner in which many functional groups react. Let’s next consider the carboxylate ion (the conjugate base of a carboxylic acid). As our example, we will use formate, the simplest possible carboxylate-containing molecule. The conjugate acid of formate is formic acid, which causes the painful sting you felt if you have ever been bitten by an ant.
Usually, you will see carboxylate groups drawn with one carbon-oxygen double bond and one carbon-oxygen single bond, with a negative formal charge located on the single-bonded oxygen. In actuality, however, the two carbon-oxygen bonds are the same length, and although there is indeed an overall negative formal charge on the group, it is shared equally between the two oxygens. Therefore, the carboxylate can be more accurately depicted by a pair of resonance contributors. Alternatively, a single structure can be used, with a dashed line depicting the resonance-delocalized pi bond and the negative charge located in between the two oxygens.
Let’s see if we can correlate these drawing conventions to a valence bond theory picture of the bonding in a carboxylate group. We know that the carbon must be sp2-hybridized, (the bond angles are close to 120˚, and the molecule is planar), and we will treat both oxygens as being sp2-hybridized as well. Both carbon-oxygen sigma bonds, then, are formed from the overlap of carbon sp2 orbitals and oxygen sp2 orbitals.
In addition, the carbon and both oxygens each have an unhybridized 2pz orbital situated perpendicular to the plane of the sigma bonds. These three 2pz orbitals are parallel to each other, and can overlap in a side-by-side fashion to form a delocalized pi bond.
Resonance contributor A shows oxygen #1 sharing a pair of electrons with carbon in a pi bond, and oxygen #2 holding a lone pair of electrons in its 2pz orbital. Resonance contributor B, on the other hand, shows oxygen #2 participating in the pi bond with carbon, and oxygen #1 holding a lone pair in its 2pz orbital. Overall, the situation is one of three parallel, overlapping 2pz orbitals sharing four delocalized pi electrons. Because there is one more electron than there are 2pz orbitals, the system has an overall charge of –1. This is the kind of 3D picture that resonance contributors are used to approximate, and once you get some practice you should be able to quickly visualize overlapping 2pz orbitals and delocalized pi electrons whenever you see resonance structures being used. In this text, carboxylate groups will usually be drawn showing only one resonance contributor for the sake of simplicity, but you should always keep in mind that the two C-O bonds are equal, and that the negative charge is delocalized to both oxygens.
Exercise 2.13: There is a third resonance contributor for formate (which we will soon learn is considered a 'minor' contributor). Draw this resonance contributor.
Here's another example, this time with a carbocation. Recall from section 2.1 that carbocations are sp2-hybridized, with an empty 2p orbital oriented perpendicular to the plane formed by three sigma bonds. If a carbocation is adjacent to a double bond, then three 2p orbitals can overlap and share the two pi electrons - another kind of conjugated pi system in which the positive charge is shared over two carbons.
Exercise 2.14: Draw the resonance contributors that correspond to the curved, two-electron movement arrows in the resonance expressions below.
Exercise 2.15: In each resonance expression, draw curved two-electron movement arrows on the left-side contributor that shows how we get to the right-side contributor. Be sure to include formal charges.
Solutions to exercises
Guided Resonance Practice
Below are a few more examples of 'legal' resonance expressions. Confirm for yourself that the octet rule is not exceeded for any atoms, and that formal charges are correct.
Exercise 2.16: Each of the 'illegal' resonance expressions below contains one or more mistakes. Explain what is incorrect in each.
Solutions to exercises
Major vs minor resonance contributors
Different resonance contributors do not always make the same contribution to the overall structure of the hybrid - rather, in many cases one contributor comes closer to depicting the actual bonding picture than another. In the case of carboxylates, contributors A and B below are equivalent in terms of their relative contribution to the hybrid structure. However, there is also a third resonance contributor ‘C, in which the carbon bears a positive formal charge and both oxygens are single-bonded and bear negative charges.
Structure C makes a less important contribution to the overall bonding picture of the group relative to A and B. How do we know that structure C is the ‘minor’ contributor? There are four basic rules which you need to learn in order to evaluate the relative importance of different resonance contributors. We will number them 5-8 so that they may be added to in the 'rules for resonance' list earlier on this page.
Rules for determining major and minor resonance contributors:
1. The carbon in contributor C does not have an octet – in general, resonance contributors in which a carbon does not fulfill the octet rule are relatively less important.
2. In structure C, a separation of charge has been introduced that is not present in A or B. In general, resonance contributors in which there is a greater separation of charge are relatively less important.
3. In structure C, there are only three bonds, compared to four in A and B. In general, a resonance structure with a lower number of total bonds is relatively less important.
4. The resonance contributor in which a negative formal charge is located on a more electronegative atom, usually oxygen or nitrogen, is more stable than one in which the negative charge is located on a less electronegative atom such as carbon. An example is in the upper left expression in the next figure.
Below are some additional examples of major and minor resonance contributors:
Why do we worry about a resonance contributor if it is the minor one? We will see later that very often a minor contributor can still be extremely important to our understanding of how a molecule reacts.
Exercise 2.17:
1. Draw a minor resonance structure for acetone (IUPAC name 2-propanone). Explain why it is a minor contributor.
2. Are acetone and 2-propanol resonance contributors of each other? Explain.
Exercise 2.18: Draw four additional resonance contributors for the molecule below. Label each one as major or minor (the structure below is of a major contributor).
Exercise 2.19: Draw three resonance contributors of methyl acetate (IUPAC name methyl methanoate), and order them according to their relative importance to the bonding picture of the molecule. Explain your reasoning.
Solutions to exercises
Resonance and peptide bonds
What is the hybridization state of the nitrogen atom in an amide? At first glance, it would seem logical to say that it is sp3-hybridized, because, like the nitrogen in an amine, the Lewis structure shows three single bonds and a lone pair. The picture looks quite different, though, if we consider another resonance contributor in which the nitrogen has a double bond to the carbonyl carbon: in this case, we would have to say that applicable hybridization is sp2, and the bonding geometry trigonal planar rather than tetrahedral.
In fact, the latter picture is more accurate: the lone pair of electrons on an amide nitrogen are not localized in an sp3 orbital, rather, they are delocalized as part of a conjugated pi system, and the bonding geometry around the nitrogen is trigonal planar as expected for sp2 hybridization. This is a good illustration of an important point: conjugation and the corresponding delocalization of electron density is stabilizing, thus if conjugation can occur, it probably will.
One of the most important examples of amide groups in nature is the ‘peptide bond’ that links amino acids to form polypeptides and proteins.
Critical to the structure of proteins is the fact that, although it is conventionally drawn as a single bond, the C-N bond in a peptide linkage has a significant barrier to rotation, indicating that to some degree, C-N pi overlap is present - in other words, there is some double bond character, and the nitrogen is sp2 hybridized with trigonal planar geometry.
The barrier to rotation in peptide bonds is an integral part of protein structure, introducing more rigidity to the protein's backbone. If there were no barrier to rotation in a peptide bond, proteins would be much more 'floppy' and three dimensional folding would be very different.
Exercise 2.20: Draw two pictures showing the unhybridized 2p orbitals and the location of pi electrons in methyl amide. One picture should represent the major resonance contributor, the other the minor contributor. How many overlapping 2p orbitals are sharing how many pi-bonded electrons?
Exercise 2.21: Draw two pictures showing the unhybridized 2p orbitals and the location of pi electrons in the 'enolate' anion shown below. One picture should represent the major resonance contributor, the other the minor contributor. How many overlapping 2p orbitals are sharing how many pi-bonded electrons?
Exercise 2.22: Below is a minor resonance contributor of a species known as an 'enamine', which we will study more in chapter 12. Draw the major resonance contributor for the enamine, and explain why your contributor is the major one (refer to resonance rules #5-8 from this section).
Solutions to exercises
Solved example: Draw the major resonance contributor of the structure below. Include in your figure the appropriate curved arrows showing how you got from the given structure to your structure. Explain why your contributor is the major one. In what kind of orbitals are the two lone pairs on the oxygen?
Solution: In the structure above, the carbon with the positive formal charge does not have a complete octet of valence electrons. Using the curved arrow convention, a lone pair on the oxygen can be moved to the adjacent bond to the left, and the electrons in the double bond shifted over to the left (see the rules for drawing resonance contributors to convince yourself that these are 'legal' moves).
The resulting resonance contributor, in which the oxygen bears the formal charge, is the major one because all atoms have a complete octet, and there is one additional bond drawn (resonance rules #5 and #7 both apply). This system can be thought of as four parallel 2p orbitals (one each on C2, C3, and C4, plus one on oxygen) sharing four pi electrons. One lone pair on the oxygen is in an unhybridized 2p orbital and is part of the conjugated pi system, and the other is located in an sp2 orbital.
Also note that one additional contributor can be drawn, but it is also minor because it has a carbon with an incomplete octet:
Exercise 2.23:
a) Draw three additional resonance contributors for the carbocation below. Include in your figure the appropriate curved arrows showing how one contributor is converted to the next.
b) Fill in the blanks: the conjugated pi system in this carbocation is composed of ______ 2p orbitals sharing ________ delocalized pi electrons.
Exercise 2.24: Draw the major resonance contributor for each of the anions below.
c) Fill in the blanks: the conjugated pi system in part (a) is composed of ______ 2p orbitals containing ________ delocalized pi electrons.
Exercise 2.25: The figure below shows how the negative formal charge on the oxygen can be delocalized to the carbon indicated by an arrow. More resonance contributors can be drawn in which negative charge is delocalized to three other atoms on the molecule.
a) Circle these atoms.
b) Draw the two most important resonance contributors for the molecule.
Solutions to exercises
A word of advice
Becoming adept at drawing resonance contributors, using the curved arrow notation to show how one contributor can be converted to another, and understanding the concepts of conjugation and resonance delocalization are some of the most challenging but also most important jobs that you will have as a beginning student of organic chemistry. If you work hard now to gain a firm grasp of these ideas, you will have come a long way toward understanding much of what follows in your organic chemistry course. Conversely, if you fail to come to grips with these concepts now, a lot of what you see later in the course will seem like a bunch of mysterious and incomprehensible lines, dots, and arrows, and you will be in for a rough ride, to say the least. More so than many other topics in organic chemistry, understanding bonding, conjugation, and resonance is something that most students really need to work on 'in person' with an instructor or tutor, preferably using a molecular modeling kit. Keep working problems, keep asking questions, and keep at it until it all makes sense!
Kahn Academy video tutorials
Drawing resonance structures
More on resonance
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/01%3A_Introduction_and_Review/1.10%3A_Resonance.txt |
LEARNING objective
• recognize acids or bases
Introduction
In 1884, the Swedish chemist Svante Arrhenius proposed two specific classifications of compounds, termed acids and bases. When dissolved in an aqueous solution, certain ions were released into the solution. As defined by Arrhenius, acid-base reactions are characterized by acids, which dissociate in aqueous solution to form hydrogen ions (H+) and bases, which form hydroxide (OH) ions. Arrhenius received the lowest passing score for his doctoral thesis with these innovative ideas about acids and bases. Ten years later he was awarded the Nobel Prize for his insights.
Acids are defined as a compound or element that releases hydrogen (H+) ions into the solution (mainly water).
$\ce{NHO_3 (aq) + H_2O(l) \rightarrow H_3O^+ + NO_3^- (aq)}$
In this reaction nitric acid (HNO3) disassociates into hydrogen (H+) and nitrate (NO3-) ions when dissolved in water. Bases are defined as a compound or element that releases hydroxide (OH-) ions into the solution.
$\ce{LiOH(s) ->[\ce{H2O}] Li^{+}(aq) + OH^{-}(aq)}$
In this reaction lithium hydroxide ($\ce{LiOH}$) dissociates into lithium ($\ce{Li^{+}}$) and hydroxide ($\ce{OH^{-}}$) ions when dissolved in water.
One way to define a class of compounds is by describing the various characteristics its members have in common. In the case of the compounds known as acids, the common characteristics include a sour taste, the ability to change the color of the vegetable dye litmus to red, and the ability to dissolve certain metals and simultaneously produce hydrogen gas. For the compounds called bases, the common characteristics are a slippery texture, a bitter taste, and the ability to change the color of litmus to blue. Acids and bases also react with each other to form compounds generally known as salts.
Although we include their tastes among the common characteristics of acids and bases, we never advocate tasting an unknown chemical!
Chemists prefer, however, to have definitions for acids and bases in chemical terms. The Swedish chemist Svante Arrhenius developed the first chemical definitions of acids and bases in the late 1800s. Arrhenius defined an acid as a compound that increases the concentration of hydrogen ion (H+) in aqueous solution. Many acids are simple compounds that release a hydrogen cation into solution when they dissolve. Similarly, Arrhenius defined a base as a compound that increases the concentration of hydroxide ion (OH) in aqueous solution. Many bases are ionic compounds that have the hydroxide ion as their anion, which is released when the base dissolves in water.
Table $1$: Formulas and Names for Some Acids and Bases
Acids Bases
Formula Name Formula Name
HCl(aq) hydrochloric acid NaOH(aq) sodium hydroxide
HBr(aq) hydrobromic acid KOH(aq) potassium hydroxide
HI(aq) hydriodic acid Mg(OH)2(aq) magnesium hydroxide
H2S(aq) hydrosulfuric acid Ca(OH)2(aq) calcium hydroxide
HC2H3O2(aq) acetic acid NH3(aq) ammonia
HNO3(aq) nitric acid
HNO2(aq) nitrous acid
H2SO4(aq) sulfuric acid
H2SO3(aq) sulfurous acid
HClO3(aq) chloric acid
HClO4(aq) perchloric acid
HClO2(aq) chlorous acid
H3PO4(aq) phosphoric acid
H3PO3(aq) phosphorous acid
Many bases and their aqueous solutions are named using the normal rules of ionic compounds that were presented previously; that is, they are named as hydroxide compounds. For example, the base sodium hydroxide (NaOH) is both an ionic compound and an aqueous solution. However, aqueous solutions of acids have their own naming rules. The names of binary acids (compounds with hydrogen and one other element in their formula) are based on the root of the name of the other element preceded by the prefix hydro- and followed by the suffix -ic acid. Thus, an aqueous solution of HCl [designated “HCl(aq)”] is called hydrochloric acid, H2S(aq) is called hydrosulfuric acid, and so forth. Acids composed of more than two elements (typically hydrogen and oxygen and some other element) have names based on the name of the other element, followed by the suffix -ic acid or -ous acid, depending on the number of oxygen atoms in the acid’s formula. Other prefixes, like per- and hypo-, also appear in the names for some acids. Unfortunately, there is no strict rule for the number of oxygen atoms that are associated with the -ic acid suffix; the names of these acids are best memorized. Table $1$ lists some acids and bases and their names. Note that acids have hydrogen written first, as if it were the cation, while most bases have the negative hydroxide ion, if it appears in the formula, written last.
The name oxygen comes from the Latin meaning “acid producer” because its discoverer, Antoine Lavoisier, thought it was the essential element in acids. Lavoisier was wrong, but it is too late to change the name now.
Example $1$
Name each substance.
1. HF(aq)
2. Sr(OH)2(aq)
Solution
1. This acid has only two elements in its formula, so its name includes the hydro- prefix. The stem of the other element’s name, fluorine, is fluor, and we must also include the -ic acid ending. Its name is hydrofluoric acid.
2. This base is named as an ionic compound between the strontium ion and the hydroxide ion: strontium hydroxide.
Exercise $1$
Name each substance.
1. H2Se(aq)
2. Ba(OH)2(aq)
Notice that one base listed in Table $1$—ammonia—does not have hydroxide as part of its formula. How does this compound increase the amount of hydroxide ion in aqueous solution? Instead of dissociating into hydroxide ions, ammonia molecules react with water molecules by taking a hydrogen ion from the water molecule to produce an ammonium ion and a hydroxide ion:
$NH_{3(aq)} + H_2O_{(ℓ)} \rightarrow NH^+_{4(aq)} + OH^−_{(aq)} \label{Eq1}$
Because this reaction of ammonia with water causes an increase in the concentration of hydroxide ions in solution, ammonia satisfies the Arrhenius definition of a base. Many other nitrogen-containing compounds are bases because they too react with water to produce hydroxide ions in aqueous solution. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/01%3A_Introduction_and_Review/1.11%3A_Arrhenius_Acids_and_Bases_%28Review%29.txt |
Learning Objective
• Use the definition of Lewis Acids and Bases to recognize electron movement in reactions
Acids and bases are an important part of chemistry. One of the most applicable theories is the Lewis acid/base motif that extends the definition of an acid and base beyond H+ and OH- ions as described by Brønsted-Lowry acids and bases.
Introduction
The Brønsted acid-base theory has been used throughout the history of acid and base chemistry. However, this theory is very restrictive and focuses primarily on acids and bases acting as proton donors and acceptors. Sometimes conditions arise where the theory doesn't necessarily fit, such as in solids and gases. In 1923, G.N. Lewis from UC Berkeley proposed an alternate theory to describe acids and bases. His theory gave a generalized explanation of acids and bases based on structure and bonding. Through the use of the Lewis definition of acids and bases, chemists are now able to predict a wider variety of acid-base reactions. Lewis' theory used electrons instead of proton transfer and specifically stated that an acid is a species that accepts an electron pair while a base donates an electron pair.
The reaction of a Lewis acid and a Lewis base will produce a coordinate covalent bond, as shown in Figure $1$ above. A coordinate covalent bond is just a type of covalent bond in which one reactant gives it electron pair to another reactant. In this case the lewis base donates its electrons to the lewis acid. When they do react this way the resulting product is called an addition compound, or more commonly an adduct.
• Lewis Acid: a species that accepts an electron pair (i.e., an electrophile) and will have vacant orbitals
• Lewis Base: a species that donates an electron pair (i.e., a nucleophile) and will have lone-pair electrons
Lewis Acids
Lewis acids accept an electron pair. Lewis Acids are Electrophilic meaning that they are electron attracting. When bonding with a base the acid uses its lowest unoccupied molecular orbital or LUMO (Figure 2).
• Various species can act as Lewis acids. All cations are Lewis acids since they are able to accept electrons. (e.g., Cu2+, Fe2+, Fe3+)
• An atom, ion, or molecule with an incomplete octet of electrons can act as an Lewis acid (e.g., BF3, AlF3).
• Molecules where the central atom can have more than 8 valence shell electrons can be electron acceptors, and thus are classified as Lewis acids (e.g., SiBr4, SiF4).
• Molecules that have multiple bonds between two atoms of different electronegativities (e.g., CO2, SO2)
Lewis Bases
Lewis Bases donate an electron pair. Lewis Bases are Nucleophilic meaning that they “attack” a positive charge with their lone pair. They utilize the highest occupied molecular orbital or HOMO (Figure 2). An atom, ion, or molecule with a lone-pair of electrons can thus be a Lewis base. Each of the following anions can "give up" their electrons to an acid, e.g., $OH^-$, $CN^-$, $CH_3COO^-$, $:NH_3$, $H_2O:$, $CO:$. Lewis base's HOMO (highest occupied molecular orbital) interacts with the Lewis acid's LUMO (lowest unoccupied molecular orbital) to create bonded molecular orbitals. Both Lewis Acids and Bases contain HOMO and LUMOs but only the HOMO is considered for Bases and only the LUMO is considered for Acids (Figure $2$).
Complex Ion / Coordination Compounds
Complex ions are polyatomic ions, which are formed from a central metal ion that has other smaller ions joined around it. While Brønsted theory can't explain this reaction Lewis acid-base theory can help. A Lewis Base is often the ligand of a coordination compound with the metal acting as the Lewis Acid (see Oxidation States of Transition Metals).
$Al^{3+} + 6 H_2O \rightleftharpoons [Al(H_2O)_6]^{3+} \label{1}$
The aluminum ion is the metal and is a cation with an unfilled valence shell, and it is a Lewis Acid. Water has lone-pair electrons and is an anion, thus it is a Lewis Base.
The Lewis Acid accepts the electrons from the Lewis Base which donates the electrons. Another case where Lewis acid-base theory can explain the resulting compound is the reaction of ammonia with Zn2+.
$Zn^{2+} + 4NH_3 \rightarrow [Zn(NH_3)_4]^{4+} \label{2}$
Similarly, the Lewis Acid is the zinc Ion and the Lewis Base is NH3. Note how Brønsted Theory of Acids and Bases will not be able to explain how this reaction occurs because there are no $H^+$ or $OH^-$ ions involved. Thus, Lewis Acid and Base Theory allows us to explain the formation of other species and complex ions which do not ordinarily contain hydronium or hydroxide ions. One is able to expand the definition of an acid and a base via the Lewis Acid and Base Theory. The lack of $H^+$ or $OH^-$ ions in many complex ions can make it harder to identify which species is an acid and which is a base. Therefore, by defining a species that donates an electron pair and a species that accepts an electron pair, the definition of a acid and base is expanded.
Amphoterism
As of now you should know that acids and bases are distinguished as two separate things however some substances can be both an acid and a base. You may have noticed this with water, which can act as both an acid or a base. This ability of water to do this makes it an amphoteric molecule. Water can act as an acid by donating its proton to the base and thus becoming its conjugate acid, OH-. However, water can also act as a base by accepting a proton from an acid to become its conjugate base, H3O+.
• Water acting as an Acid:
$H_2O + NH_3 \rightarrow NH_4^+ + OH^- \label{3}$
• Water acting as a Base:
$H_2O + HCl \rightarrow Cl^- + H_3O^+ \label{4}$
You may have noticed that the degree to which a molecule acts depends on the medium in which the molecule has been placed in. Water does not act as an acid in an acid medium and does not act as a base in a basic medium. Thus, the medium which a molecule is placed in has an effect on the properties of that molecule. Other molecules can also act as either an acid or a base. For example,
$Al(OH)_3 + 3H^+ \rightarrow Al^{3+} + 3H_2O \label{5}$
• where Al(OH)3 is acting as a Lewis Base.
$Al(OH)_3 + OH^- \rightarrow Al(OH)_4^- \label{6}$
• where Al(OH)3 is acting as an Lewis Acid.
Note how the amphoteric properties of the Al(OH)3 depends on what type of environment that molecule has been placed in.
Lewis Bases & Acids as Nucleophiles & Electrophiles
The emphasis on electron flow in the Lewis Theory of acids and bases is an important foundation for learning and predicting reaction mechanisms. The electron rich Lewis base can be described as a nucleophile. Nucleophiles are attracted to and can react with compounds or ions that have full or partial positive charge (like the nucleus). The electron poor Lewis acids can be described as electrophiles. Electrophiles attract nucleophiles until orbital overlap occurs between them triggering a reaction. At this point in the course, we can indicate electron flow using curved arrows when both the reactant(s) and product(s) are given.
Example
Exercises
For the following reactions,
a) add curved arrows to indicate the electron flow
b) label each reactant as the Nu (nucleophile) or E+ (electrophile).
Outside Links
• Very Detailed review of Lewis Acids and Bases, covering all topics of this type of chemistry
• Very Complex and Detailed "Lewis Acid and Base Interaction Matrix"
• Youtube Video about Lewis Acids/Bases
Contributors and Attributions
• Adam Abudra (UCD), Tajinder Badial (UCD) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/01%3A_Introduction_and_Review/1.12%3A_Lewis_Acids_and_Bases.txt |
Learning Objective
• Determine relative strengths of acids and bases from their pKa values
• Determine the form of an acid or base at a specified pH (given the pKa)
The Henderson-Hasselbach Equation - a Quantitative View
We will use the general reaction for a weak acid to write the Ka expression.
\[HA_{(aq)} + H_2O_{(l)} \rightleftharpoons H_3O^+_{(aq)} + A^-_{(aq)}\]
\[K_a = \dfrac{[H_3O^+][A^-]}{[HA]}\]
pKa = -log Ka
where each bracketed term represents the concentration of that substance in solution.
The stronger an acid, the greater the ionization, the lower the pKa, and the lower the pH the compound will produce in solution.
It is important to realize that pKa is not at all the same thing as pH: the former is an inherent property of a compound or functional group, while the latter is the measure of the hydronium ion concentration in a particular aqueous solution:
pH = -log [H3O+]
Additional reagents can be added to a reaction solution to change the pH of the reaction conditions beyond the effects of an individual compound.
Relative Acidity and pKa Values
An application of the Henderson-Hasselbach Equation is the ability to determine the relative acidity of compounds by comparing their pKa values. The stronger an acid, the greater the ionization, the lower the pKa, and the lower the pH the compound will produce in solution. Some selected pKa values for compounds in the study of organic chemistry are shown bellow. Since organic reactions can be performed in non-aqueous environments, the pH can exceed 14 and organic compounds can have pKa values above 16. It is a variation on that line from the Wizard of Oz, "We don't live in water anymore."
It is a very good idea to commit to memory the approximate pKa ranges of the compounds above. A word of caution: when using the pKa table, be absolutely sure that you are considering the correct conjugate acid/base pair. If you are asked to say something about the basicity of ammonia (NH3) compared to that of ethoxide ion (CH3CH2O-), for example, the relevant pKa values to consider are 9.2 (the pKa of ammonium ion) and 16 (the pKa of ethanol). From these numbers, you know that ethoxide is the stronger base. Do not make the mistake of using the pKa value of 38: this is the pKa of ammonia acting as an acid, and tells you how basic the NH2- ion is (very basic!)
* A note on the pKa of water: The pKa of water is 14. Biochemistry and organic chemistry texts often list the value as 15.7. These texts have incorrectly factored the molar value for the concentration of water into the equilibrium constant. The correct derivation of the equilibrium constant involves the activity of water, which has a value of 1.
Example
While this course begins with single functional groups, we will eventually work with interesting compounds containing multiple functional groups. Recognizing which hydrogens can be ionized as acidic protons and which hydrogens can NOT, is a useful skill. Notice in this example that we need to evaluate the potential acidity at four different locations on the molecule.
Aldehyde and aromatic protons are not at all acidic (pKavalues are above 40 – not on our table). The two protons on the carbon next to the carbonyl are slightly acidic, with pKa values around 19-20 according to the table. The most acidic proton is on the phenol group, so if the compound were to be subjected to a single molar equivalent of strong base, this is the proton that would be donated.
Acidic & Basic Environments - Everything is Relative in Reactivity
Because our goal is understanding dynamic chemical reactivity, we do NOT need to know the specific amount of the protonated and unprotonated forms of a compound. We simply need to know which form is predominate. When the pH of the environment is less than the pKa of the compound, the environment is considered acidic and the compound will exist predominately in its protonated form. When the pH of the environment is greater than the pKa of the compound, the environment is considered basic and the compound will exist predominately in its deprotonated form.
For example, the pKa of acetic acid is about 5. At a pH of 1, the environment is considered acidic and acetic acid exists predominately in its protonated form. At pH 8, the environment is considered basic, and acetic acid becomes deprotonated to form acetate (CH3CO2-). Conversely, the pKa of phenol is 10. At pH 8, the environment is considered acidic for phenol and it remains primarily protonated.
It is also important to remember that organic chemistry does NOT have to occur in water so pKa values can be as high as 50.
Exercise
1. Complete the table below to indicate whether each compound exists predominantly in its protonated (acidic environment) or deprotonated (basic environment) form.
compound (pKa) pH 1 environment pH 8 environment pH 13 environment
Answer
1.
compound (pKa) pH 1 environment pH 8 environment pH 13 environment | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/01%3A_Introduction_and_Review/1.13%3A_Distinguishing_between_pH_and_pKa.txt |
Learning Objective
• Predict relative strengths of acids and bases from their structure, bonding and resonance
Since compounds are neutral, it can be difficult to evaluate and compare their overall stability without going through the tedious process of performing bond energy calculations.
When acidic compounds donate hydrogen ions or accepts electrons, they become ionized. It is much easier to compare ions because we can evaluate the charge density. The lower the charge density, the more stable the ion. Conversely, the higher the charge density, the less stable the ion. Charge density is analogous to density of matter. We place charge in the numerator, instead of mass, and volume can still be found in the denominator.
$\text{charge density} =\dfrac{\text{charge}}{\text{volume}}$
The six strong acids (HCl, HBr, HI, HNO3, H2SO4, HClO4) fully ionize to form the highly stable anions (Cl-, Br-, I-, NO3-, SO42-, ClO4-) respectively.
For the remaining weak acids (HA), we can determine their relative acidity by comparing the relative electron densities of their conjugate bases (A-).
the lower the electron density, the more stable the conjugate base
Structural Effects on Electron Density - Four Considerations
There are four main considerations for evaluating electron density.
1. Identity of the element or atoms holding the charge
2. Can the charge be delocalized by resonance?
3. Are there any inductive effects?
4. Hybridization of orbital holding the charge
These considerations are listed in order of importance and are explained individually, but must be looked at collectively.
Identity of the Element
When comparing the identity of the elements, it depends on the positional relationship of the elements on the periodic table.
Within a Group (aka down a column) As we move down the periodic table, the electrons are occupying higher energy subshells creating a larger atomic size and volume. As the volume increases, the electron density decreases.
Figure $1$ shows spheres representing the atoms of the s and p blocks from the periodic table to scale, showing the two trends for the atomic radius.
This relationship of atomic size and electron density is illustrated when we compare the relative acidities of methanol, CH3OH, with methanethiol, CH3SH. The lower pKa value of 10.4 for methanethiol indicates that it is a stronger acid than methanol with a pKa value of 15.5. It is important to remember that neither compound is considered an acid. These relationships become useful when trying to deprotonate compounds to increase their chemical reactivity in non-aqueous reaction conditions.
Across a Period (aka across a row) As we move across a period of the main group elements, the valence electrons all occupy orbitals in the same shell. These electrons have comparable energy, so this factor does not help us discern differences relative stability. Differences in electronegativity are now the dominant factor. This trend is shown when comparing the pKa values of methane, ammonia, water, and hydrofluoric acid reflects the relative electonegativities of the C < N < O < F.
Compound pKa Reaction
methane 50 $CH_4 \rightleftharpoons CH_3^- + H^+$
ammonia 36 $NH_3 \rightleftharpoons NH_2^- + H^+$
water 14 $H_2O \rightleftharpoons OH^- + H^+$
hydrofluoric acid 3 $HF \rightleftharpoons F^- + H^+$
Periodic trends
First, we will focus on individual atoms, and think about trends associated with the position of an element on the periodic table. We’ll use as our first models the simple organic compounds ethane, methylamine, and ethanol, but the concepts apply equally to more complex biomolecules, such as the side chains of alanine, lysine, and serine.
We can see a clear trend in acidity as we move from left to right along the second row of the periodic table from carbon to nitrogen to oxygen. The key to understanding this trend is to consider the hypothetical conjugate base in each case: the more stable (weaker) the conjugate base, the stronger the acid. Look at where the negative charge ends up in each conjugate base. In the ethyl anion, the negative charge is borne by carbon, while in the methylamine anion and ethoxide anion the charges are located on a nitrogen and an oxygen, respectively. Remember the periodic trend in electronegativity (section 2.3A): it also increases as we move from left to right along a row, meaning that oxygen is the most electronegative of the three, and carbon the least. The more electronegative an atom, the better it is able to bear a negative charge. Thus, the ethoxide anion is the most stab = 0.00le (lowest energy, least basic) of the three conjugate bases, and the ethyl anion is the least stable (highest energy, most basic).
We can use the same set of ideas to explain the difference in basicity between water and ammonia.
$pK_a = 0.00 \,\,\,\,\,\, H_3O^+ \rightleftharpoons H_2O + H^+$
$pK_a = 9.26 \,\,\,\,\,\, NH_4^+ \rightleftharpoons NH_3 + H^+$
By looking at the pKavalues for the appropriate conjugate acids, we know that ammonia is more basic than water. Oxygen, as the more electronegative element, holds more tightly to its lone pair than the nitrogen. The nitrogen lone pair, therefore, is more likely to break away and form a new bond to a proton - it is, in other words, more basic. Once again, a more reactive (stronger) conjugate base means a less reactive (weaker) conjugate acid.
When moving vertically within a given column of the periodic table, we again observe a clear periodic trend in acidity. This is best illustrated with the halides: basicity, like electronegativity, increases as we move up the column.
Conversely, acidity in the haloacids increases as we move down the column.
In order to make sense of this trend, we will once again consider the stability of the conjugate bases. Because fluorine is the most electronegative halogen element, we might expect fluoride to also be the least basic halogen ion. But in fact, it is the least stable, and the most basic! It turns out that when moving vertically in the periodic table, the size of the atom trumps its electronegativity with regard to basicity. The atomic radius of iodine is approximately twice that of fluorine, so in an iodine ion, the negative charge is spread out over a significantly larger volume:
This illustrates a fundamental concept in organic chemistry that is important enough to put in red:
Electrostatic charges, whether positive or negative, are more stable when they are ‘spread out’ than when they are confined to one atom.
We will see this idea expressed again and again throughout our study of organic reactivity, in many different contexts. For now, the concept is applied only to the influence of atomic radius on anion stability. Because fluoride is the least stable (most basic) of the halide conjugate bases, HF is the least acidic of the haloacids, only slightly stronger than acetic acid. HI, with a pKa of about -9, is one the strongest acids known.
More importantly to the study of biological organic chemistry, this trend tells us that thiols are more acidic than alcohols. The pKa of the thiol group on the cysteine side chain, for example, is approximately 8.3, while the pKa for the hydroxyl on the serine side chain is on the order of 17.
To reiterate: acid strength increases as we move to the right along a row of the periodic table, and as we move down a column.
Example $1$:
Draw the structure of the conjugate base that would form if the compound below were to react with 1 molar equivalent of sodium hydroxide:
Solution
Is resonance possible to localize the charge?
In the previous section we focused our attention on periodic trends - the differences in acidity and basicity between groups where the exchangeable proton was bound to different elements. Now, it is time to think about how the structure of different organic groups contributes to their relative acidity or basicity, even when we are talking about the same element acting as the proton donor/acceptor. When evaluating conjugate bases for the presence of the resonance contributors, remember to look for movable electrons as described in section 1.10 of this chapter. Delocalizing electrons over two or more atoms lowers the electron density.
The first model pair we will consider is ethanol and acetic acid, but the conclusions we reach will be equally valid for all alcohol and carboxylic acid groups. Despite the fact that they are both oxygen acids, the pKa values of ethanol and acetic acid are very different. What makes a carboxylic acid so much more acidic than an alcohol? As before, we begin by considering the conjugate bases.
In both species, the negative charge on the conjugate base is held by an oxygen, so periodic trends cannot be invoked. For acetic acid, however, there is a key difference: a resonance contributor can be drawn in which the negative charge is localized on the second oxygen of the group. The two resonance forms for the conjugate base are equal in energy, according to our ‘rules of resonance’ (section 2.2C). What this means, you may recall, is that the negative charge on the acetate ion is not located on one oxygen or the other: rather it is shared between the two. Chemists use the term ‘delocalization of charge’ to describe this situation. In the ethoxide ion, by contrast, the negative charge is ‘locked’ on the single oxygen – it has nowhere else to go.
Now is the time to think back to that statement from the previous section that was so important that it got printed in bold font in its own paragraph – in fact, it is so important that we’ll just say it again: "Electrostatic charges, whether positive or negative, are more stable when they are ‘spread out’ than when they are confined to one atom." Now, we are seeing this concept in another context, where a charge is being ‘spread out’ (in other words, delocalized) by resonance, rather than simply by the size of the atom involved.
The delocalization of charge by resonance has a very powerful effect on the reactivity of organic molecules, enough to account for the difference of over 12 pKa units between ethanol and acetic acid (and remember, pKa is a log expression, so we are talking about a difference of over 1012 between the acidity constants for the two molecules). The acetate ion is that much more stable than the ethoxide ion, all due to the effects of resonance delocalization.
The resonance effect also nicely explains why a nitrogen atom is basic when it is in an amine, but not basic when it is part of an amide group. Recall that in an amide, there is significant double-bond character to the carbon-nitrogen bond, due to a second resonance contributor in which the nitrogen lone pair is part of a p bond.
While the electron lone pair of an amine nitrogen is ‘stuck’ in one place, the lone pair on an amide nitrogen is delocalized by resonance. Notice that in this case, we are extending our central statement to say that electron density – in the form of a lone pair – is stabilized by resonance delocalization, even though there is not a negative charge involved. Here’s another way to think about it: the lone pair on an amide nitrogen is not available for bonding with a proton – these two electrons are too ‘comfortable’ being part of the delocalized pi-bonding system. The lone pair on an amine nitrogen, by contrast, is not part of a delocalized p system, and is very ready to form a bond with any acidic proton that might be nearby.
Often it requires some careful thought to predict the most acidic proton on a molecule. Ascorbic acid, also known as Vitamin C, has a pKa of 4.1.
There are four hydroxyl groups on this molecule – which one is the most acidic? If we consider all four possible conjugate bases, we find that there is only one for which we can delocalized the negative charge over two oxygen atoms.
Example $1$:
Rank the compounds below from most acidic to least acidic, and explain your reasoning.
(CC-NC-SA; Timothy Soderberg via UMn Morris Digital Well)
Inductive Effects
The inductive effect is an experimentally observed effect of the transmission of charge through a chain of atoms in a molecule, resulting in a permanent dipole in a bond. Inductive effects decrease quickly with distance. The inductive effect can be electron donating which helps stabilize positive charge. Alkyl groups (hydrocarbons) are inductive electron donators. The inductive effect can also be electron withdrawing. Electronegativity indicates the strength of electron withdrawing induction. Halogens are inductive electron withdrawing groups.
The effects of induction on relative acidity can also be seen when comparing acetic acid with trifluoroacetic acid. The difference in acidity does not have to do with resonance delocalization because no additional resonance structures can be drawn for the fluorinated molecule. The fluorine atoms inductively pull some of the electron density away from the carboxylate ion to further delocalize the negative charge of the conjugate base.
Compare the pKa values of acetic acid and its mono-, di-, and tri-chlorinated derivatives:
(CC-NC-SA; Timothy Soderberg via UMn Morris Digital Well)
The presence of the chlorines clearly increases the acidity of the carboxylic acid group. A chlorine atom is more electronegative than a hydrogen, and thus is able to ‘induce’, or ‘pull’ electron density towards itself, away from the carboxylate group. In helping to further spread out the electron density of the conjugate base, which as we know has a stabilizing effect. In this context, the chlorine substituent is called an electron-withdrawing group. Notice that the pKa-lowering effect of each chlorine atom, while significant, is not as dramatic as the delocalizing resonance effect illustrated by the difference in pKa values between an alcohol and a carboxylic acid. In general, resonance effects are more powerful than inductive effects. The inductive electron-withdrawing effect of the chlorines takes place through covalent bonds, and its influence decreases markedly with distance – thus a chlorine two carbons away from a carboxylic acid group has a decreased effect compared to a chlorine just one carbon away.
Exercise
Rank the compounds below from most acidic to least acidic, and explain your reasoning.
Solution
Orbital Hybridization
The hybridization of an orbital affects its electronegativity. Within a shell, the s orbitals occupy the region closer to the nucleus than the p orbitals. Therefore, the spherical s orbitals are more electronegative than the lobed p orbitals. The relative electronegativity of hybridized orbitals is sp > sp2 > sp3. This trend indicates the sp hybridized orbitals are more stable with a -1 charge than sp3 hybridized orbitals. The table below shows how orbital hybridization compares with the identity of the atom when predicting relative acidity.
Guided Practice
Let's practice by comparing the relative acidity of phenol with acetic acid.
Which compound is the stronger acid - phenol or acetic acid?
To answer this question, we draw all the relevant resonance contributors for each conjugate base, phenoxide and acetate, respectively.
Phenoxide has four resonance contributors, but three of the contributors have a negative charge on a carbon atom while both resonance contributors for acetate have a negative charge on the more electronegative element oxygen. There are no inductive effects or orbital hybridization differences to consider in this example, so we would predict acetic acid to be the stronger acid. The acetate ion is more stable than the phenoxide ion, so we would expect acetic acid to be the stronger acid.
The pKa table below supports our prediction. Acetic acid has a pKa of 4.7 while phenol has a pKa of 9.9.
A word of caution: when using the pKa table, be absolutely sure that you are considering the correct conjugate acid/base pair. If you are asked to say something about the basicity of ammonia (NH3) compared to that of ethoxide ion (CH3CH2O-), for example, the relevant pKa values to consider are 9.2 (the pKa of ammonium ion) and 16 (the pKa of ethanol). From these numbers, you know that ethoxide is the stronger base. Do not make the mistake of using the pKa value of 38: this is the pKa of ammonia acting as an acid, and tells you how basic the NH2- ion is (very basic!)
Contributors and Attributions
• Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris)
• Layne A. Morsch (University of Illinois Springfield) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/01%3A_Introduction_and_Review/1.14%3A_Predicting_Relative_Acidity.txt |
Learning objective
• Determine the empirical and molecular formulas from combustion data
Molecular formulas tell you how many atoms of each element are in a compound, and empirical formulas tell you the simplest or most reduced ratio of elements in a compound. If a compound's molecular formula cannot be reduced any more, then the empirical formula is the same as the molecular formula. Combustion analysis can determine the empirical formula of a compound, but cannot determine the molecular formula (other techniques can though). Once known, the molecular formula can be calculated from the empirical formula.
Empirical Formulas
An empirical formula tells us the relative ratios of different atoms in a compound. The ratios hold true on the molar level as well. Thus, H2O is composed of two atoms of hydrogen and 1 atom of oxygen. Likewise, 1.0 mole of H2O is composed of 2.0 moles of hydrogen and 1.0 mole of oxygen. We can also work backwards from molar ratios since if we know the molar amounts of each element in a compound we can determine the empirical formula.
Example $1$: Mercury Chloride
Mercury forms a compound with chlorine that is 73.9% mercury and 26.1% chlorine by mass. What is the empirical formula?
Let's say we had a 100 gram sample of this compound. The sample would therefore contain 73.9 grams of mercury and 26.1 grams of chlorine. How many moles of each atom do the individual masses represent?
For Mercury:
$(73.9 \;g) \times \left(\dfrac{1\; mol}{200.59\; g}\right) = 0.368 \;moles$
For Chlorine:
$(26.1\; g) \times \left(\dfrac{1\; mol}{35.45\; g}\right) = 0.736\; mol$
What is the molar ratio between the two elements?
$\dfrac{0.736 \;mol \;Cl}{0.368\; mol\; Hg} = 2.0$
Thus, we have twice as many moles (i.e. atoms) of Cl as Hg. The empirical formula would thus be (remember to list cation first, anion last):
$HgCl_2$
Molecular Formula from Empirical Formula
The chemical formula for a compound obtained by composition analysis is always the empirical formula. We can obtain the chemical formula from the empirical formula if we know the molecular weight of the compound. The chemical formula will always be some integer multiple of the empirical formula (i.e. integer multiples of the subscripts of the empirical formula). The general flow for this approach is shown in Figure $1$ and demonstrated in Example $2$.
Example $2$: Ascorbic Acid
Vitamin C (ascorbic acid) contains 40.92 % C, 4.58 % H, and 54.50 % O, by mass. The experimentally determined molecular mass is 176 amu. What is the empirical and chemical formula for ascorbic acid?
Solution
Consider an arbitrary amount of 100 grams of ascorbic acid, so we would have:
• 40.92 grams C
• 4.58 grams H
• 54.50 grams O
This would give us how many moles of each element?
• Carbon
$(40.92\; \cancel{g\; C}) \times \left( \dfrac{1\; mol\; C}{12.011\; \cancel{g\; C}} \right) = 3.407\; mol \; C$
• Hydrogen
$(4.58\; \cancel{g\; H}) \times \left( \dfrac{1\; mol\; H}{1.008\; \cancel{g\; H}} \right) = 4.544\; mol \;H$
• Oxygen
$(54.50\; \cancel{g\; O}) \times \left( \dfrac{1\; mol\; O}{15.999\; \cancel{g\; O}} \right) = 3.406\; mol \; O$
Determine the simplest whole number ratio by dividing by the smallest molar amount (3.406 moles in this case - see oxygen):
• Carbon
$C= \dfrac{3.407\; mol}{3.406\; mol} \approx 1.0$
• Hydrogen
$C= \dfrac{4.5.44\; mol}{3.406\; mol} = 1.0$
• Oxygen
$C= \dfrac{3.406\; mol}{3.406\; mol} = 1.0$
The relative molar amounts of carbon and oxygen appear to be equal, but the relative molar amount of hydrogen is higher. Since we cannot have "fractional" atoms in a compound, we need to normalize the relative amount of hydrogen to be equal to an integer. 1.333 would appear to be 1 and 1/3, so if we multiply the relative amounts of each atom by '3', we should be able to get integer values for each atom.
C = (1.0)*3 = 3
H = (1.333)*3 = 4
O = (1.0)*3 = 3
or
C3H4O3
This is our empirical formula for ascorbic acid.
What about the chemical formula? We are told that the experimentally determined molecular mass is 176 amu. What is the molecular mass of our empirical formula?
(3*12.011) + (4*1.008) + (3*15.999) = 88.062 amu
The molecular mass from our empirical formula is significantly lower than the experimentally determined value. What is the ratio between the two values?
(176 amu/88.062 amu) = 2.0
Thus, it would appear that our empirical formula is essentially one half the mass of the actual molecular mass. If we multiplied our empirical formula by '2', then the molecular mass would be correct. Thus, the actual molecular formula is:
2* C3H4O3 = C6H8O6
Combustion Analysis
When a compound containing carbon and hydrogen is subject to combustion with oxygen in a special combustion apparatus all the carbon is converted to CO2 and the hydrogen to H2O (Figure $2$). The amount of carbon produced can be determined by measuring the amount of CO2 produced. This is trapped by the sodium hydroxide, and thus we can monitor the mass of CO2 produced by determining the increase in mass of the CO2 trap. Likewise, we can determine the amount of H produced by the amount of H2O trapped by the magnesium perchlorate.
One of the most common ways to determine the elemental composition of an unknown hydrocarbon is an analytical procedure called combustion analysis. A small, carefully weighed sample of an unknown compound that may contain carbon, hydrogen, nitrogen, and/or sulfur is burned in an oxygen atmosphere,Other elements, such as metals, can be determined by other methods. and the quantities of the resulting gaseous products (CO2, H2O, N2, and SO2, respectively) are determined by one of several possible methods. One procedure used in combustion analysis is outlined schematically in Figure $3$ and a typical combustion analysis is illustrated in Examples $3$ and $4$.
Example $3$: Combustion of Isopropyl Alcohol
What is the empirical formulate for isopropyl alcohol (which contains only C, H and O) if the combustion of a 0.255 grams isopropyl alcohol sample produces 0.561 grams of CO2 and 0.306 grams of H2O?
Solution
From this information quantitate the amount of C and H in the sample.
$(0.561\; \cancel{g\; CO_2}) \left( \dfrac{1 \;mol\; CO_2}{44.0\; \cancel{g\;CO_2}}\right)=0.0128\; mol \; CO_2$
Since one mole of CO2 is made up of one mole of C and two moles of O, if we have 0.0128 moles of CO2 in our sample, then we know we have 0.0128 moles of C in the sample. How many grams of C is this?
$(0.0128 \; \cancel{mol\; C}) \left( \dfrac{12.011\; g \; C}{1\; \cancel{mol\;C}}\right)=0.154\; g \; C$
How about the hydrogen?
$(0.306 \; \cancel{g\; H_2O}) \left( \dfrac{1\; mol \; H_2O}{18.0\; \cancel{g \;H_2O}}\right)=0.017\; mol \; H_2O$
Since one mole of H2O is made up of one mole of oxygen and two moles of hydrogen, if we have 0.017 moles of H2O, then we have 2*(0.017) = 0.034 moles of hydrogen. Since hydrogen is about 1 gram/mole, we must have 0.034 grams of hydrogen in our original sample.
When we add our carbon and hydrogen together we get:
0.154 grams (C) + 0.034 grams (H) = 0.188 grams
But we know we combusted 0.255 grams of isopropyl alcohol. The 'missing' mass must be from the oxygen atoms in the isopropyl alcohol:
0.255 grams - 0.188 grams = 0.067 grams oxygen
This much oxygen is how many moles?
$(0.067 \; \cancel{g\; O}) \left( \dfrac{1\; mol \; O}{15.994\; \cancel{g \;O}}\right)=0.0042\; mol \; O$
Overall therefore, we have:
• 0.0128 moles Carbon
• 0.0340 moles Hydrogen
• 0.0042 moles Oxygen
Divide by the smallest molar amount to normalize:
• C = 3.05 atoms
• H = 8.1 atoms
• O = 1 atom
Within experimental error, the most likely empirical formula for propanol would be $C_3H_8O$
Example $4$: Combustion of Naphalene
Naphthalene, the active ingredient in one variety of mothballs, is an organic compound that contains carbon and hydrogen only. Complete combustion of a 20.10 mg sample of naphthalene in oxygen yielded 69.00 mg of CO2 and 11.30 mg of H2O. Determine the empirical formula of naphthalene.
Given: mass of sample and mass of combustion products
Asked for: empirical formula
Strategy:
1. Use the masses and molar masses of the combustion products, CO2 and H2O, to calculate the masses of carbon and hydrogen present in the original sample of naphthalene.
2. Use those masses and the molar masses of the elements to calculate the empirical formula of naphthalene.
Solution:
A Upon combustion, 1 mol of $\ce{CO2}$ is produced for each mole of carbon atoms in the original sample. Similarly, 1 mol of H2O is produced for every 2 mol of hydrogen atoms present in the sample. The masses of carbon and hydrogen in the original sample can be calculated from these ratios, the masses of CO2 and H2O, and their molar masses. Because the units of molar mass are grams per mole, we must first convert the masses from milligrams to grams:
$mass \, of \, C = 69.00 \, mg \, CO_2 \times {1 \, g \over 1000 \, mg } \times {1 \, mol \, CO_2 \over 44.010 \, g \, CO_2} \times {1 \, mol C \over 1 \, mol \, CO_2 } \times {12.011 \,g \over 1 \, mol \, C}$
$= 1.883 \times 10^{-2} \, g \, C$
$mass \, of \, H = 11.30 \, mg \, H_2O \times {1 \, g \over 1000 \, mg } \times {1 \, mol \, H_2O \over 18.015 \, g \, H_2O} \times {2 \, mol H \over 1 \, mol \, H_2O } \times {1.0079 \,g \over 1 \, mol \, H}$
$= 1.264 \times 10^{-3} \, g \, H$
B To obtain the relative numbers of atoms of both elements present, we need to calculate the number of moles of each and divide by the number of moles of the element present in the smallest amount:
$moles \, C = 1.883 \times 10^{-2} \,g \, C \times {1 \, mol \, C \over 12.011 \, g \, C} = 1.568 \times 10^{-3} \, mol C$
$moles \, H = 1.264 \times 10^{-3} \,g \, H \times {1 \, mol \, H \over 1.0079 \, g \, H} = 1.254 \times 10^{-3} \, mol H$
Dividing each number by the number of moles of the element present in the smaller amount gives
$H: {1.254\times 10^{−3} \over 1.254 \times 10^{−3}} = 1.000 \, \, \, C: {1.568 \times 10^{−3} \over 1.254 \times 10^{−3}}= 1.250$
Thus naphthalene contains a 1.25:1 ratio of moles of carbon to moles of hydrogen: C1.25H1.0. Because the ratios of the elements in the empirical formula must be expressed as small whole numbers, multiply both subscripts by 4, which gives C5H4 as the empirical formula of naphthalene. In fact, the molecular formula of naphthalene is C10H8, which is consistent with our results.
Exercise 1 $4$
1. Xylene, an organic compound that is a major component of many gasoline blends, contains carbon and hydrogen only. Complete combustion of a 17.12 mg sample of xylene in oxygen yielded 56.77 mg of CO2 and 14.53 mg of H2O. Determine the empirical formula of xylene.
2. The empirical formula of benzene is CH (its molecular formula is C6H6). If 10.00 mg of benzene is subjected to combustion analysis, what mass of CO2 and H2O will be produced?
Answer a
The empirical formula is C4H5. (The molecular formula of xylene is actually C8H10.)
Answer b
33.81 mg of CO2; 6.92 mg of H2O
Exercise 2
Elemental analysis of an organic compound indicates its composition to be 37.82% carbon, 6.36% hydrogen, and 55.82% chlorine.
a. What is the empirical formula for this compound?
b. Mass spectral analysis indicates a molar mass of 129 g/mol. What is the molecular formula for this compound?
c. Draw all the possible bond-line structures with this molecular formula.
Solutions to Exercise 2
a. C2H5Cl with a molar mass of 64.5 g/mol
b. C4H10Cl2
c. There 8 possible structures with the molecular formula C4H10Cl2. It can help to start with the different carbon backbones and then systematically add any branches (substituents). | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/01%3A_Introduction_and_Review/1.15%3A_Molecular_Formulas_and_Empirical_Formulas_%28Review%29.txt |
Bond Formation: The Octet Rule
1-1 Identify the number of valence electrons for each of the following elements. Then, identify the maximum number of covalent bonds it can form with other atoms while keeping a neutral net charge.
a) Oxygen
b) Carbon
c) Chlorine
d) Sulfur
e) Hydrogen
f) Boron
1-2 Which of the following atoms can bond with Br - to satisfy the octet rule?
a) Mg+2
b) O-2
c) Cl-
d) K+
1-3 Draw the Lewis dot structure of the correct answer from the previous problem 1-2 (a) - (d).
1-4 Identify which of the following compounds could not form due to an unfilled octet.
a) NCl3
b) NaOH
c) PCl
d) CF4
Lewis Structures
1-5 Draw the Lewis structures for the following compounds.
a) H2O
b) O3
c) BH3
d) SOCl2
1-6 Name the element that corresponds to each electronic configuration and identify how many valence electrons it has.
a) 1s22s22p6
b) 1s22s22p63s2
c) 1s22s22p4
d) 1s22s22p63s23p64s23d104p5
1-7 Draw the Lewis structures for PF3 and PF5.
1-8 Draw the Lewis structure for furan.
1-9 Identify the correct Lewis structure for hydroperoxyl, HO2.
Electronegativity and Bond Polarity
1-10 For the indicated bond in each of the following compounds, identify which atom is more electronegative, if applicable.
1-11 For each of the compounds in the previous problem, add a dipole moment arrow.
Formal Charges
1-12 For the following compounds, draw the structural formula. Then calculate the formal charge on each atom other than hydrogen.
a) N(CH3)4+
b) HSO4-
c) CH3CC-
1-13 Identify the formal charge for the following compounds.
1-14 Identify the formal charges for the central carbon in each of the following compounds.
Ionic Structures
1-15 Identify the substituent ions that make up the following salts.
a) NaCl
b) MgBr2
c) KNO3
d) NaH2PO4
1-16 Identify the products of the following reactions.
1-17 Give the correct nomenclature or write the correct chemical formula for the following ionic compounds.
a) NaCN
b) calcium oxalate
c) Al(OH)3
d) tin (II) phosphate
e) potassium hypochlorite
Resonance
1-18 For the following structure, draw its resonance structure(s).
1-19 Which resonance form from the previous problem has the most stable carbocation? Explain your answer.
1-20 Draw the important resonance forms to show the delocalization of charges in the following compounds.
1-21 Explain how resonance contributes to the lower pKa of acetic acid CH3CO2H (pKa= 4.75) compared to the pKa of ethanol CH3CH2OH (pKa=15.9).
1-22 Draw the resonance structure(s) for fulminic acid (HCNO).
Structural, Molecular and Empirical Formulas
1-23 Identify the molecular and empirical formula for the following structures.
1-24 Draw all possible structural formulas for the following compounds.
a) C4H10
b) CHN
c) C4H9Cl
1-25 True or False: You can always calculate the exact molecular weight of a molecule from its empirical formula.
1-26 For the following molecular formulas, provide the empirical formula.
a) C4H4O2
b) C8H6N2
c) C9H21N3O3
Acids and Bases - Arrhenius, Bronsted-Lowry, and Lewis
1-27 Briefly explain the three different definitions of acids and bases.
1-28 Calculate the Ka of nitric acid (HNO3). pKa of nitric acid is -1.4.
1-29 Rank the following in order of decreasing acidity: NH4+ HF H3O+ H2O
1-30 Rank the following in order of decreasing basicity: HSO4- H2O CH3COO- NH2-
1-31 Identify which compound is the stronger base. Identify which compound is the stronger acid.
1.32 Identify which group is more likely to grab a H+.
1.17: Solutions to Additional Exercises
Bond Formation: The Octet Rule
1-1:
a) 6 v.e. / 2 covalent bonds
b) 4 v.e. / 4 covalent bonds
c) 7 v.e. / 1 covalent bond
d) 6 v.e. / 2 covalent bonds (can also expand the octet to make 6 covalent bonds)
e) 1 v.e. / 1 covalent bond
f) 3 v.e. / 3 covalent bonds
1-2:
(d) K +
1-3:
1-4:
(c) PCl
Lewis Structures
1-5:
1-6:
a) Neon, 8 valence electrons
b) Magnesium, 2 valence electrons
c) Oxygen, 6 valence electrons
d) Bromine, 7 valence electrons
1-7:
1-8:
1-9: C.
Electronegativity and Bond Polarity
1-10:
1-11:
a) no dipole moment arrow b/c non-polar
1-12:
1-13:
a) 0
b) -1
c) -2
1-14:
a) 0
b) -1
c) 0
d) +1
Ionic Structures
1-15:
a) Na+ and Cl-
b) Mg+2 and 2 Br-
c) K+ and NO3-
d) Na+ and H2PO4-
1-16:
1-17:
a) sodium cyanide
b) CaC2O4
c) aluminum hydroxide
d) Sn3(PO4)2
e) KClO
Resonance
1-18:
1-19:
The resonance structure that is most stable has the tertiary carbocation. This tertiary carbocation is stabilized by hyperconjugation as well as two possible directions for resonance (compared to one immediate resonance structure for the other two carbocations).
1-20:
1-21:
When comparing the deprotonated forms of acetic acid and ethanol, acetate and ethoxide respectively, you can observe that acetate delocalizes the negative charge over the entire carboxylate group. Ethoxide, however, can only hold the negative charge on the alkoxide, making it a better base, but worse as an acid.
1-22:
Structural, Molecular and Empirical Formulas
1-23:
1-24:
1-25:
False; empirical formulas are the simplest whole number ratios that are useful in calculating percent compositions of atoms in a molecule. However, as they do not give the absolute number of atoms in a molecule, they cannot be used to calculate the molecular weight of the molecule.
1-26:
a) C2H2O
b) C4H3N
c) C9H7NO
Acids and Bases - Arrhenius, Bronsted-Lowry, and Lewis
1-27:
Arrhenius: An Arrhenius acid is a species that will donate a H+ when dissolved in water. An Arrhenius base is a species that will break down to yield a OH- when dissolved in water.
Bronsted-Lowry: A Bronsted-Lowry acid is a species that will donate a H+ when dissolved in solution (not only in water). A Bronsted-Lowry base is a species that can accept a H+ in solution (not only in water).
Lewis: A Lewis acid is an electron pair acceptor. A Lewis base is an electron base donor.
1-28:
Ka = 2.4 x 101
1-29:
H3O+ > HF > NH4+ > H2O
1-30:
NH2- > H2O > CH3COO- > HSO4-
1-31:
Compound A is the stronger base. Compound B is the stronger acid.
1-32:
Group A will want to grab the H+ more than group B. Since C is less electronegative than N, it cannot stabilize the negative charge as well and will want to grab a H+ in order to get rid of the charge. (grab = react with) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/01%3A_Introduction_and_Review/1.16%3A_Additional_Exercises.txt |
learning objective
• recognize acids and bases
The Brønsted-Lowry Theory of Acids and Bases
In 1923, Danish chemist Johannes Brønsted and English chemist Thomas Lowry independently proposed new definitions for acids and bases, ones that focus on proton transfer. A Brønsted-Lowry acid is any species that can donate a proton (H+) to another molecule. A Brønsted-Lowry base is any species that can accept a proton from another molecule. In short, a Brønsted-Lowry acid is a proton donor (PD), while a Brønsted-Lowry base is a proton acceptor (PA).
A Brønsted-Lowry acid is a proton donor, while a Brønsted-Lowry base is a proton acceptor.
Let us use the reaction of ammonia in water to demonstrate the Brønsted-Lowry definitions of an acid and a base. Ammonia and water molecules are reactants, while the ammonium ion and the hydroxide ion are products:
$\ce{NH3(aq) + H2O (ℓ) <=> NH^{+}4(aq) + OH^{−}(aq) }\label{Eq1}$
What has happened in this reaction is that the original water molecule has donated a hydrogen ion to the original ammonia molecule, which in turn has accepted the hydrogen ion. We can illustrate this as follows:
Because the water molecule donates a hydrogen ion to the ammonia, it is the Brønsted-Lowry acid, while the ammonia molecule—which accepts the hydrogen ion—is the Brønsted-Lowry base. Thus, ammonia acts as a base in both the Arrhenius sense and the Brønsted-Lowry sense.
Is an Arrhenius acid like hydrochloric acid still an acid in the Brønsted-Lowry sense? Yes, but it requires us to understand what really happens when HCl is dissolved in water. Recall that the hydrogen atom is a single proton surrounded by a single electron. To make the hydrogen ion, we remove the electron, leaving a bare proton. Do we really have bare protons floating around in aqueous solution? No, we do not. What really happens is that the H+ ion attaches itself to H2O to make H3O+, which is called the hydronium ion. For most purposes, H+ and H3O+ represent the same species, but writing H3O+ instead of H+ shows that we understand that there are no bare protons floating around in solution. Rather, these protons are actually attached to solvent molecules.
The Hydronium IOn
A proton in aqueous solution may be surrounded by more than one water molecule, leading to formulas like $\ce{H5O2^{+}}$ or $\ce{H9O4^{+}}$ rather than $\ce{H3O^{+}}$. It is simpler, however, to use $\ce{H3O^{+}}$ to represent the hydronium ion.
With this in mind, how do we define HCl as an acid in the Brønsted-Lowry sense? Consider what happens when HCl is dissolved in H2O:
$\ce{HCl(g) + H_2O (ℓ) \rightarrow H_3O^{+}(aq) + Cl^{−}(aq) }\label{Eq2}$
We can depict this process using Lewis electron dot diagrams:
Now we see that a hydrogen ion is transferred from the HCl molecule to the H2O molecule to make chloride ions and hydronium ions. As the hydrogen ion donor, HCl acts as a Brønsted-Lowry acid; as a hydrogen ion acceptor, H2O is a Brønsted-Lowry base. So HCl is an acid not just in the Arrhenius sense but also in the Brønsted-Lowry sense. Moreover, by the Brønsted-Lowry definitions, H2O is a base in the formation of aqueous HCl. So the Brønsted-Lowry definitions of an acid and a base classify the dissolving of HCl in water as a reaction between an acid and a base—although the Arrhenius definition would not have labeled H2O a base in this circumstance.
• A Brønsted-Lowry acid is a proton (hydrogen ion) donor.
• A Brønsted-Lowry base is a proton (hydrogen ion) acceptor.
• All Arrhenius acids and bases are Brønsted-Lowry acids and bases as well. However, not all Brønsted-Lowry acids and bases are Arrhenius acids and bases.
Example $1$
Aniline (C6H5NH2) is slightly soluble in water. It has a nitrogen atom that can accept a hydrogen ion from a water molecule just like the nitrogen atom in ammonia does. Write the chemical equation for this reaction and identify the Brønsted-Lowry acid and base.
Solution
C6H5NH2 and H2O are the reactants. When C6H5NH2 accepts a proton from H2O, it gains an extra H and a positive charge and leaves an OH ion behind. The reaction is as follows:
$\ce{C6H5NH2(aq) + H2O(ℓ) <=> C6H5NH3^{+}(aq) + OH^{−}(aq)} \nonumber$
Because C6H5NH2 accepts a proton, it is the Brønsted-Lowry base. The H2O molecule, because it donates a proton, is the Brønsted-Lowry acid.
Exercise $1$
Identify the Brønsted-Lowry acid and the Brønsted-Lowry base in this chemical equation.
$\ce{H2PO4^{-} + H_2O <=> HPO4^{2-} + H3O^{+}}$
Answer:
Brønsted-Lowry acid: H2PO4-; Brønsted-Lowry base: H2O
Exercise $2$
Which of the following compounds is a Bronsted-Lowry base?
1. HCl
2. HPO42-
3. H3PO4
4. NH4+
5. CH3NH3+
Answer:
A Brønsted-Lowry Base is a proton acceptor, which means it will take in an H+. This eliminates $\ce{HCl}$, $\ce{H3PO4}$ , $\ce{NH4^{+}}$ and $\ce{CH_3NH_3^{+}}$ because they are Bronsted-Lowry acids. They all give away protons. In the case of $\ce{HPO4^{2-}}$, consider the following equation:
$\ce{HPO4^{2-} (aq) + H2O (l) \rightarrow PO4^{3-} (aq) + H3O^{+}(aq) } \nonumber$
Here, it is clear that HPO42- is the acid since it donates a proton to water to make H3O+ and PO43-. Now consider the following equation:
$\ce{ HPO4^{2-}(aq) + H2O(l) \rightarrow H2PO4^{-} + OH^{-}(aq)} \nonumber$
In this case, HPO42- is the base since it accepts a proton from water to form H2PO4- and OH-. Thus, HPO42- is an acid and base together, making it amphoteric.
Since HPO42- is the only compound from the options that can act as a base, the answer is (b) HPO42-.
Conjugate Acid-Base Pair
In reality, all acid-base reactions involve the transfer of protons between acids and bases. For example, consider the acid-base reaction that takes place when ammonia is dissolved in water. A water molecule (functioning as an acid) transfers a proton to an ammonia molecule (functioning as a base), yielding the conjugate base of water, $\ce{OH^-}$, and the conjugate acid of ammonia, $\ce{NH4+}$:
In the reaction of ammonia with water to give ammonium ions and hydroxide ions, ammonia acts as a base by accepting a proton from a water molecule, which in this case means that water is acting as an acid. In the reverse reaction, an ammonium ion acts as an acid by donating a proton to a hydroxide ion, and the hydroxide ion acts as a base. The conjugate acid–base pairs for this reaction are $NH_4^+/NH_3$ and $H_2O/OH^−$.
The strongest acids are at the bottom left, and the strongest bases are at the top right. The conjugate base of a strong acid is a very weak base, and, conversely, the conjugate acid of a strong base is a very weak acid.
Example $2$
Identify the conjugate acid-base pairs in this equilibrium.
$\ce{CH3CO2H + H2O <=> H3O^{+} + CH3CO2^{-}} \nonumber$
Solution
Similarly, in the reaction of acetic acid with water, acetic acid donates a proton to water, which acts as the base. In the reverse reaction, $H_3O^+$ is the acid that donates a proton to the acetate ion, which acts as the base.
Once again, we have two conjugate acid–base pairs:
• the parent acid and its conjugate base ($CH_3CO_2H/CH_3CO_2^−$) and
• the parent base and its conjugate acid ($H_3O^+/H_2O$).
Example $3$
Identify the conjugate acid-base pairs in this equilibrium.
$(CH_{3})_{3}N + H_{2}O\rightleftharpoons (CH_{3})_{3}NH^{+} + OH^{-} \nonumber$
Solution
One pair is H2O and OH, where H2O has one more H+ and is the conjugate acid, while OH has one less H+ and is the conjugate base.
The other pair consists of (CH3)3N and (CH3)3NH+, where (CH3)3NH+ is the conjugate acid (it has an additional proton) and (CH3)3N is the conjugate base.
Exercise $3$
Identify the conjugate acid-base pairs in this equilibrium.
$\ce{NH2^{-} + H2O\rightleftharpoons NH3 + OH^{-}} \nonumber$
Answer:
H2O (acid) and OH (base); NH2 (base) and NH3 (acid)
Contributors and Attributions
• Henry Agnew (UC Davis) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/01%3A_Introduction_and_Review/1.18%3A_Brnsted-Lowry_Acids_and_Bases_%28Review%29.txt |
Learning Objectives
After reading this chapter and completing ALL the exercises, a student can be able to
• define the terms "sterics" and "electrostatics" - refer to section 2.1
• write and interpret molecular orbital (MO) diagrams - refer to section 2.2
• predict the hybridization and geometry of atoms in a molecule - refer to section 2.3
• draw accurate 3-D representations of molecules with approximate bond angles - refer to section 2.3
• recognize conjugated pi bond systems - refer to section 2.4
• recognize that benzene is aromatic - refer to section 2.4
• identify the orbitals occupied by lone pair electrons - refer to section 2.5
• distinguish between bonds that can rotate and those that cannot - refer to section 2.6
• recognize the relationships between constitutional (structural) isomers, conformational isomers, and geometric isomers - refer to section 2.7
• apply the homologous series to organic molecules with 1-10 carbons - refer to section 2.8
• classify hydrocarbons as saturated or unsaturated - refer to section 2.8
• classify hydrocarbons as alkanes, alkenes, alkynes, cycloalkanes, or aromatics (arenes) - refer to section 2.8
• recognize and classify the common functional groups of organic chemistry (alkanes, alkenes, alkynes, alkyl halides, alcohols, amines, ethers, aldehydes, ketones, carboxylic acids, esters, and amides - refer to section 2.9
• determine the dominant intermolecular forces (IMFs) of organic compounds - refer to section 2.10
• predict the relative boil points of organic compounds - refer to section 2.11
• predict whether a mixture of compounds will a form homogeneous or heterogeneous solution - refer to section 2.12
• distinguish between organic compounds that are H-bond donors versus H-bond acceptors - refer to section 2.13
• apply the terms sterics and electrostatics to organic compounds - refer to sections 2.1- 2.13
• 2.1: Pearls of Wisdom
A few "pearls of wisdom" about "sterics" and "electrostatics" to provide context when applying the concepts of general chemistry to organic compounds.
• 2.2: Molecular Orbital (MO) Theory (Review)
Molecular orbital (MO) theory describes the behavior of electrons in a molecule in terms of combinations of the atomic wavefunctions.
• 2.3: Hybridization and Molecular Shapes (Review)
Hybridization and the Valence Shell Electron Pair Repulsion Theory effectively predict the three-dimensional structure of organic molecules. Since carbon can only form four bonds, we can limit our study to the tetrahedral, trigonal planar, and linear electron geometries.
• 2.4: 2.4 Conjugated Pi Bond Systems
A conjugated system is a system of connected p-orbitals with delocalized electrons in compounds with alternating single and multiple bonds, which in general may lower the overall energy of the molecule and increase stability. Recognizing the conjugated systems is helpful in determining reaction pathways.
• 2.5: Lone Pair Electrons and Bonding Theories
The chemical reactivity of lone pair electrons can be determined from the identity of the orbital they occupy. This concept will be further refined when we study aromaticity.
• 2.6: Bond Rotation
Single bonds can rotate, while double and triple bonds are rigid.
• 2.7: Isomerism Introduction
Structural (constitutional) isomers have the same molecular formula but a different bonding arrangement among the atoms. Stereoisomers have identical molecular formulas and arrangements of atoms. They differ from each other only in the spatial orientation of groups in the molecule.
• 2.8: Hydrocarbons and the Homologous Series
Since we will be spending the rest of the course working with compounds with a carbon backbone, there is no time like the present to learn the homologous series, the names for simple, straight hydrocarbon chains and branches.
• 2.9: Organic Functional Groups
Functional groups are to organic chemistry what ions are to general chemistry. We simply must be able to recognize and distinguish between functional grouops to learn organic chemistry.
• 2.10: Intermolecular Forces (IMFs) - Review
Intermolecular forces (IMFs) have many useful applications in organic chemistry. For students interested in biochemistry, the concepts of IMFs are called non-covalent interactions when they occur within a large biological molecule creating secondary and tertiary structure.
• 2.11: Intermolecular Forces and Relative Boiling Points (bp)
The relative strength of the intermolecular forces (IMFs) can be used to predict the relative boiling points of pure substances.
• 2.12: Intermolecular Forces and Solubilities
Organic chemistry can perform reactions in non-aqueous solutions using organic solvents. It is important to start considering the solvent as a reaction parameter and the solubility of each reagent.
• 2.13: Additional Practice Problems
• 2.14: Organic Functional Groups- H-bond donors and H-bond acceptors
When evaluating organic compounds, we want to visualize the compounds in their three-dimensional shapes exerting intermolecular forces on their environment. Because reactions will occur in aqueous and non-aqueous (organic) solutions, it is important to recognize which functional groups are both H-bond donors and H-bond acceptors and which groups are only H-bond acceptors.
• 2.15: Solutions to Additional Exercises
This section has the solutions to the exercises in the previous section.
• 2.16: Additional Exercises
This section has additional exercises for the key learning objectives of this chapter.
02: Structure and Properties of Organic Molecules
Learning Objective
• define the terms "sterics" and "electrostatics"
Introduction
Functional groups are the common bonding patterns found in organic compounds. Organic compounds are classified by their functional groups.
To talk about organic chemistry, we need to be able to
a) recognize and name the major organic functional groups (see chapter 3 for nomenclature)
b) apply bonding theories to the structure of functional groups
c) visual functional groups in three dimensions
d) determine the polarity & intermolecular forces of organic compounds
Ultimately, all of the information above will be integrated at the end of this chapter to predict solubilities and relative boiling points of organic compounds. In future chapters, these skills will help elucidate reaction mechanisms and pathways.
Sterics & Electrostatics - all roads lead to one or other
Sterics and electrostatics are primary considerations when learning the reactions of organic chemistry.
Sterics is the spatial arrangement (3-dimensional structure) of atoms in a molecule or ion.
Electrostatics is the the charge distribution within a molecule or ion.
Depending on the reaction mechanism, either sterics or electrostatics (charge stabilization) will play a dominant role in the rate determining step. For concerted (one-step) reactions, sterics will strongly influence the orientation of reactants in the transition state. For two-step reactions, there is typically a charged intermediate that requires stabilization for the reaction to proceed. The intermediate with the lowest charge distribution is the most stable and reacts preferentially.
Sterics can be predicted using bonding theories. Molecular orbital (MO) theory uses the combination of atomic orbitals to yield molecular orbitals that are delocalized over the entire molecule. In valence bond theory (VB) theory, atomic orbitals can be hybridized. VB theory assumes that all bonds are localized bonds formed between two atoms by the donation of an electron from each atom. As discussed in chapter 1, this assumption is invalid because some atoms can bond using delocalized electrons through resonance. VB theory does a good job of qualitatively describing the shapes of covalent compounds which is important in determining the sterics of the reactions. While Molecular Orbital (MO) theory is good for understanding bonding in general and the electrostatics of a reactant, intermediate, or product.
Electrostatics are determined by applying the same concepts used to determine the relative acidity of compounds by evaluating the electron density of their conjugate bases. The less charge, the more stable an ion is. The stability of ions is determined by the identity of the element being ionized, charge delocalization via resonance, inductive effects, and orbital hybridization. Inductive effects can be electron withdrawing (aka electronegative) or electron donating, such alkyl group stabilization of carbocations. These parameters are listed in order of importance with the overall character of an ion must be evaluated to determine its relative stability. Refer to section 1.15 of this text for the full explanation. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/02%3A_Structure_and_Properties_of_Organic_Molecules/2.01%3A_Pearls_of_Wisdom.txt |
Learning Objective
• write and interpret molecular orbital (MO) diagrams
Overview
Molecular orbital (MO) theory describes the behavior of electrons in a molecule in terms of combinations of the atomic wavefunctions. The resulting molecular orbitals may extend over all the atoms in the molecule. Bonding molecular orbitals are formed by in-phase combinations of atomic wavefunctions, and electrons in these orbitals stabilize a molecule. Antibonding molecular orbitals result from out-of-phase combinations and electrons in these orbitals make a molecule less stable.
Molecular orbital theory describes the distribution of electrons in molecules in much the same way that the distribution of electrons in atoms is described using atomic orbitals. Using quantum mechanics, the behavior of an electron in a molecule is still described by a wave function, Ψ, analogous to the behavior in an atom. Just like electrons around isolated atoms, electrons around atoms in molecules are limited to discrete (quantized) energies. The region of space in which a valence electron in a molecule is likely to be found is called a molecular orbital (Ψ2). Like an atomic orbital, a molecular orbital is full when it contains two electrons with opposite spin.
We will consider the molecular orbitals in molecules composed of two identical atoms (H2 or Cl2, for example). Such molecules are called homonuclear diatomic molecules. In these diatomic molecules, several types of molecular orbitals occur.
The mathematical process of combining atomic orbitals to generate molecular orbitals is called the linear combination of atomic orbitals (LCAO). The wave function describes the wavelike properties of an electron. Molecular orbitals are combinations of atomic orbital wave functions. Combining waves can lead to constructive interference, in which peaks line up with peaks, or destructive interference, in which peaks line up with troughs (Figure $2$). In orbitals, the waves are three dimensional, and they combine with in-phase waves producing regions with a higher probability of electron density and out-of-phase waves producing nodes, or regions of no electron density.
There are two types of molecular orbitals that can form from the overlap of two atomic s orbitals on adjacent atoms. The two types are illustrated in Figure 8.4.3. The in-phase combination produces a lower energy σs molecular orbital (read as "sigma-s") in which most of the electron density is directly between the nuclei. The out-of-phase addition (which can also be thought of as subtracting the wave functions) produces a higher energy $σ^∗_s$ molecular orbital (read as "sigma-s-star") molecular orbital in which there is a node between the nuclei. The asterisk signifies that the orbital is an antibonding orbital. Electrons in a σs orbital are attracted by both nuclei at the same time and are more stable (of lower energy) than they would be in the isolated atoms. Adding electrons to these orbitals creates a force that holds the two nuclei together, so we call these orbitals bonding orbitals. Electrons in the $σ^∗_s$ orbitals are located well away from the region between the two nuclei. The attractive force between the nuclei and these electrons pulls the two nuclei apart. Hence, these orbitals are called antibonding orbitals. Electrons fill the lower-energy bonding orbital before the higher-energy antibonding orbital, just as they fill lower-energy atomic orbitals before they fill higher-energy atomic orbitals.
In p orbitals, the wave function gives rise to two lobes with opposite phases, analogous to how a two-dimensional wave has both parts above and below the average. We indicate the phases by shading the orbital lobes different colors. When orbital lobes of the same phase overlap, constructive wave interference increases the electron density. When regions of opposite phase overlap, the destructive wave interference decreases electron density and creates nodes. When p orbitals overlap end to end, they create σ and σ* orbitals (Figure $4$). If two atoms are located along the x-axis in a Cartesian coordinate system, the two px orbitals overlap end to end and form σpx (bonding) and $σ^∗_{px}$ (antibonding) (read as "sigma-p-x" and "sigma-p-x star," respectively). Just as with s-orbital overlap, the asterisk indicates the orbital with a node between the nuclei, which is a higher-energy, antibonding orbital.
The side-by-side overlap of two p orbitals gives rise to a pi (π) bonding molecular orbital and a π* antibonding molecular orbital, as shown in Figure $5$. In valence bond theory, we describe π bonds as containing a nodal plane containing the internuclear axis and perpendicular to the lobes of the p orbitals, with electron density on either side of the node. In molecular orbital theory, we describe the π orbital by this same shape, and a π bond exists when this orbital contains electrons. Electrons in this orbital interact with both nuclei and help hold the two atoms together, making it a bonding orbital. For the out-of-phase combination, there are two nodal planes created, one along the internuclear axis and a perpendicular one between the nuclei.
In the molecular orbitals of diatomic molecules, each atom also has two sets of p orbitals oriented side by side (py and pz), so these four atomic orbitals combine pairwise to create two π orbitals and two π* orbitals. The πpy and $π^∗_{py}$ orbitals are oriented at right angles to the πpz and $π^∗_{pz}$ orbitals. Except for their orientation, the πpy and πpz orbitals are identical and have the same energy; they are degenerate orbitals. The $π^∗_{py}$ and $π^∗_{pz}$ antibonding orbitals are also degenerate and identical except for their orientation. A total of six molecular orbitals results from the combination of the six atomic p orbitals in two atoms: σpx and $σ^∗_{px}$, πpy and $π^∗_{py}$, πpz and $π^∗_{pz}$.
Example $1$
Molecular Orbitals Predict what type (if any) of molecular orbital would result from adding the wave functions so each pair of orbitals shown overlap. The orbitals are all similar in energy.
Solution
1. This is an in-phase combination, resulting in a σ3p orbital
2. This will not result in a new orbital because the in-phase component (bottom) and out-of-phase component (top) cancel out. Only orbitals with the correct alignment can combine.
3. This is an out-of-phase combination, resulting in a $π^∗_{3p}$ orbital.
Exercise $1$
Label the molecular orbital shown as σ or π, bonding or antibonding and indicate where the node occurs.
Answer
The orbital is located along the internuclear axis, so it is a σ orbital. There is a node bisecting the internuclear axis, so it is an antibonding orbital.
Molecular Orbital Energy Diagrams
The relative energy levels of atomic and molecular orbitals are typically shown in a molecular orbital diagram (Figure $7$). For a diatomic molecule, the atomic orbitals of one atom are shown on the left, and those of the other atom are shown on the right. Each horizontal line represents one orbital that can hold two electrons. The molecular orbitals formed by the combination of the atomic orbitals are shown in the center. Dashed lines show which of the atomic orbitals combine to form the molecular orbitals. For each pair of atomic orbitals that combine, one lower-energy (bonding) molecular orbital and one higher-energy (antibonding) orbital result. Thus we can see that combining the six 2p atomic orbitals results in three bonding orbitals (one σ and two π) and three antibonding orbitals (one σ* and two π*).
We predict the distribution of electrons in these molecular orbitals by filling the orbitals in the same way that we fill atomic orbitals, by the Aufbau principle. Lower-energy orbitals fill first, electrons spread out among degenerate orbitals before pairing, and each orbital can hold a maximum of two electrons with opposite spins (Figure $7$). Just as we write electron configurations for atoms, we can write the molecular electronic configuration by listing the orbitals with superscripts indicating the number of electrons present. For clarity, we place parentheses around molecular orbitals with the same energy. In this case, each orbital is at a different energy, so parentheses separate each orbital. Thus we would expect a diatomic molecule or ion containing seven electrons (such as $\ce{Be2+}$) would have the molecular electron configuration $(σ_{1s})^2(σ^∗_{1s})^2(σ_{2s})^2(σ^∗_{2s})^1$. It is common to omit the core electrons from molecular orbital diagrams and configurations and include only the valence electrons.
Bond Order
The filled molecular orbital diagram shows the number of electrons in both bonding and antibonding molecular orbitals. The net contribution of the electrons to the bond strength of a molecule is identified by determining the bond order that results from the filling of the molecular orbitals by electrons.
When using Lewis structures to describe the distribution of electrons in molecules, we define bond order as the number of bonding pairs of electrons between two atoms. Thus a single bond has a bond order of 1, a double bond has a bond order of 2, and a triple bond has a bond order of 3. We define bond order differently when we use the molecular orbital description of the distribution of electrons, but the resulting bond order is usually the same. The MO technique is more accurate and can handle cases when the Lewis structure method fails, but both methods describe the same phenomenon.
In the molecular orbital model, an electron contributes to a bonding interaction if it occupies a bonding orbital and it contributes to an antibonding interaction if it occupies an antibonding orbital. The bond order is calculated by subtracting the destabilizing (antibonding) electrons from the stabilizing (bonding) electrons. Since a bond consists of two electrons, we divide by two to get the bond order. We can determine bond order with the following equation:
$\textrm{bond order}=\dfrac{(\textrm{number of bonding electrons})−(\textrm{number of antibonding electrons})}{2}$
The order of a covalent bond is a guide to its strength; a bond between two given atoms becomes stronger as the bond order increases. If the distribution of electrons in the molecular orbitals between two atoms is such that the resulting bond would have a bond order of zero, a stable bond does not form. We next look at some specific examples of MO diagrams and bond orders.
Bonding in Diatomic Molecules
A dihydrogen molecule (H2) forms from two hydrogen atoms. When the atomic orbitals of the two atoms combine, the electrons occupy the molecular orbital of lowest energy, the σ1s bonding orbital. A dihydrogen molecule, H2, readily forms because the energy of a H2 molecule is lower than that of two H atoms. The σ1s orbital that contains both electrons is lower in energy than either of the two 1s atomic orbitals.
A molecular orbital can hold two electrons, so both electrons in the H2 molecule are in the σ1s bonding orbital; the electron configuration is $(σ_{1s})^2$. We represent this configuration by a molecular orbital energy diagram (Figure $8$) in which a single upward arrow indicates one electron in an orbital, and two (upward and downward) arrows indicate two electrons of opposite spin.
A dihydrogen molecule contains two bonding electrons and no antibonding electrons so we have
$\ce{bond\: order\: in\: H2}=\dfrac{(2−0)}{2}=1$
Because the bond order for the H–H bond is equal to 1, the bond is a single bond.
A helium atom has two electrons, both of which are in its 1s orbital. Two helium atoms do not combine to form a dihelium molecule, He2, with four electrons, because the stabilizing effect of the two electrons in the lower-energy bonding orbital would be offset by the destabilizing effect of the two electrons in the higher-energy antibonding molecular orbital. We would write the hypothetical electron configuration of He2 as $(σ_{1s})^2(σ^∗_{1s})^2$ as in Figure $9$ . The net energy change would be zero, so there is no driving force for helium atoms to form the diatomic molecule. In fact, helium exists as discrete atoms rather than as diatomic molecules. The bond order in a hypothetical dihelium molecule would be zero.
$\ce{bond\: order\: in\: He2}=\dfrac{(2−2)}{2}=0$
A bond order of zero indicates that no bond is formed between two atoms.
C2 | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/02%3A_Structure_and_Properties_of_Organic_Molecules/2.02%3A_Molecular_Orbital_%28MO%29_Theory_%28Review%29.txt |
Learning Objectives
• predict the hybridization and geometry of atoms in a molecule - refer to section 2.3
• draw accurate 3-D representations of molecules with approximate bond angles
Formation of sigma bonds: the H2 molecule
The simplest case to consider is the hydrogen molecule, H2. When we say that the two electrons from each of the hydrogen atoms are shared to form a covalent bond between the two atoms, what we mean in valence bond theory terms is that the two spherical 1s orbitals overlap, allowing the two electrons to form a pair within the two overlapping orbitals.
These two electrons are now attracted to the positive charge of both of the hydrogen nuclei, with the result that they serve as a sort of ‘chemical glue’ holding the two nuclei together.
Bonding in Methane
Now let’s turn to methane, the simplest organic molecule. Recall the valence electron configuration of the central carbon:
This picture, however, is problematic. How does the carbon form four bonds if it has only two half-filled p orbitals available for bonding? A hint comes from the experimental observation that the four C-H bonds in methane are arranged with tetrahedral geometry about the central carbon, and that each bond has the same length and strength. In order to explain this observation, valence bond theory relies on a concept called orbital hybridization. In this picture, the four valence orbitals of the carbon (one 2s and three 2p orbitals) combine mathematically (remember: orbitals are described by equations) to form four equivalent hybrid orbitals, which are named sp3 orbitals because they are formed from mixing one s and three p orbitals. In the new electron configuration, each of the four valence electrons on the carbon occupies a single sp3 orbital.
The sp3 hybrid orbitals, like the p orbitals of which they are partially composed, are oblong in shape, and have two lobes of opposite sign. Unlike the p orbitals, however, the two lobes are of very different size. The larger lobes of the sp3 hybrids are directed towards the four corners of a tetrahedron, meaning that the angle between any two orbitals is 109.5o.
This geometric arrangement makes perfect sense if you consider that it is precisely this angle that allows the four orbitals (and the electrons in them) to be as far apart from each other as possible.This is simply a restatement of the Valence Shell Electron Pair Repulsion (VSEPR) theory that you learned in General Chemistry: electron pairs (in orbitals) will arrange themselves in such a way as to remain as far apart as possible, due to negative-negative electrostatic repulsion.
Each C-H bond in methane, then, can be described as an overlap between a half-filled 1s orbital in a hydrogen atom and the larger lobe of one of the four half-filled sp3 hybrid orbitals in the central carbon. The length of the carbon-hydrogen bonds in methane is 1.09 Å (1.09 x 10-10 m).
While previously we drew a Lewis structure of methane in two dimensions using lines to denote each covalent bond, we can now draw a more accurate structure in three dimensions, showing the tetrahedral bonding geometry. To do this on a two-dimensional page, though, we need to introduce a new drawing convention: the solid / dashed wedge system. In this convention, a solid wedge simply represents a bond that is meant to be pictured emerging from the plane of the page. A dashed wedge represents a bond that is meant to be pictured pointing into, or behind, the plane of the page. Normal lines imply bonds that lie in the plane of the page.
This system takes a little bit of getting used to, but with practice your eye will learn to immediately ‘see’ the third dimension being depicted.
Example
Imagine that you could distinguish between the four hydrogens in a methane molecule, and labeled them Ha through Hd. In the images below, the exact same methane molecule is rotated and flipped in various positions. Draw the missing hydrogen atom labels. (It will be much easier to do this if you make a model.)
Exercise
Describe, with a picture and with words, the bonding in chloroform, CHCl3.
Solutions
The bonding arrangement here is also tetrahedral: the three N-H bonds of ammonia can be pictured as forming the base of a trigonal pyramid, with the fourth orbital, containing the lone pair, forming the top of the pyramid.
Recall from your study of VSEPR theory in General Chemistry that the lone pair, with its slightly greater repulsive effect, ‘pushes’ the three N-H sbonds away from the top of the pyramid, meaning that the H-N-H bond angles are slightly less than tetrahedral, at 107.3˚ rather than 109.5˚.
VSEPR theory also predicts, accurately, that a water molecule is ‘bent’ at an angle of approximately 104.5˚. It would seem logical, then, to describe the bonding in water as occurring through the overlap of sp3-hybrid orbitals on oxygen with 1sorbitals on the two hydrogen atoms. In this model, the two nonbonding lone pairs on oxygen would be located in sp3 orbitals.
Some experimental evidence, however, suggests that the bonding orbitals on the oxygen are actually unhybridized 2p orbitals rather than sp3 hybrids. Although this would seem to imply that the H-O-H bond angle should be 90˚ (remember that p orbitals are oriented perpendicular to one another), it appears that electrostatic repulsion has the effect of distorting this p-orbital angle to 104.5˚. Both the hybrid orbital and the nonhybrid orbital models present reasonable explanations for the observed bonding arrangement in water, so we will not concern ourselves any further with the distinction.
Exercise
Draw, in the same style as the figures above, an orbital picture for the bonding in methylamine.
Solution
Formation of $\pi$ bonds - $sp^2$ and $sp$ hybridization
The valence bond theory, along with the hybrid orbital concept, does a very good job of describing double-bonded compounds such as ethene. Three experimentally observable characteristics of the ethene molecule need to be accounted for by a bonding model:
1. Ethene is a planar (flat) molecule.
2. Bond angles in ethene are approximately 120o, and the carbon-carbon bond length is 1.34 Å, significantly shorter than the 1.54 Å single carbon-carbon bond in ethane.
3. There is a significant barrier to rotation about the carbon-carbon double bond.
Clearly, these characteristics are not consistent with an sp3 hybrid bonding picture for the two carbon atoms. Instead, the bonding in ethene is described by a model involving the participation of a different kind of hybrid orbital. Three atomic orbitals on each carbon – the 2s, 2px and 2py orbitals – combine to form three sp2 hybrids, leaving the 2pz orbital unhybridized.
The three sp2 hybrids are arranged with trigonal planar geometry, pointing to the three corners of an equilateral triangle, with angles of 120°between them. The unhybridized 2pz orbital is perpendicular to this plane (in the next several figures, sp2 orbitals and the sigma bonds to which they contribute are represented by lines and wedges; only the 2pz orbitals are shown in the 'space-filling' mode).
The carbon-carbon double bond in ethene consists of one sbond, formed by the overlap of two sp2 orbitals, and a second bond, calleda π (pi) bond, which is formed by the side-by-side overlap of the two unhybridized 2pz orbitals from each carbon.
spacefilling image of bonding in ethene
The pi bond does not have symmetrical symmetry. Because they are the result of side-by-side overlap (rather then end-to-end overlap like a sigma bond), pi bonds are not free to rotate. If rotation about this bond were to occur, it would involve disrupting the side-by-side overlap between the two 2pz orbitals that make up the pi bond. The presence of the pi bond thus ‘locks’ the six atoms of ethene into the same plane. This argument extends to larger alkene groups: in each case, the six atoms of the group form a single plane.
Conversely, sbonds such as the carbon-carbon single bond in ethane (CH3CH3) exhibit free rotation, and can assume many different conformations, or shapes - this is one of the main subjects of Chapter 3.
Exercise
Circle the six atoms in the molecule below that are ‘locked’ into the same plane.
Exercise
What kinds of orbitals are overlapping in bonds a-d indicated below?
Exercise
What is wrong with the way the following structure is drawn?
Solutions
A similar picture can be drawn for the bonding in carbonyl groups, such as formaldehyde. In this molecule, the carbon is sp2-hybridized, and we will assume that the oxygen atom is also sp2 hybridized. The carbon has three sigma bonds: two are formed by overlap between two of its sp2 orbitals with the 1sorbital from each of the hydrogens, and the third sigma bond is formed by overlap between the remaining carbon sp2 orbital and an sp2 orbital on the oxygen. The two lone pairs on oxygen occupy its other two sp2 orbitals.
The pi bond is formed by side-by-side overlap of the unhybridized 2pz orbitals on the carbon and the oxygen. Just like in alkenes, the 2pz orbitals that form the pi bond are perpendicular to the plane formed by the sigma bonds.
Exercise
Describe and draw the bonding picture for the imine group shown below. Use the drawing of formaldehyde above as your guide.
Solution | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/02%3A_Structure_and_Properties_of_Organic_Molecules/2.03%3A_Hybridization_and_Molecular_Shapes_%28Review%29.txt |
Learning Objective
• recognize conjugated pi bond systems
• recognize that benzene is aromatic
Introduction
It is important to train our eye to recognize structural features that have stabilizing effects. Alternating single and double bonds create a conjugated pi bond system across multiple atoms that lowers the energy and stabilizes the molecule or ion. When we look at carbon-carbon double bonds (C=C), we need to look and see if they are isolated or conjugated.
To understand the source of this stabilization we will use molecular orbital (MO) theory. Valence bond theory does a remarkably good job at explaining the bonding geometry of many of the functional groups in organic compounds, however, it fails to adequately account for the stability contained in alternating double and single bonds. In order to understand these properties, we will use the ideas of MO theory.
Let’s go back and consider again the simplest possible covalent bond: the one in molecular hydrogen (H2). When we described the hydrogen molecule using valence bond theory, we said that the two 1s orbitals from each atom overlap, allowing the two electrons to be shared and thus forming a covalent bond. In molecular orbital theory, we make a further statement: we say that the two atomic 1s orbitals mathematically combine to form two new orbitals. Recall that an atomic orbital (such as the 1s orbital of a hydrogen atom) describes a region of space around a single atom inside which electrons are likely to be found. A molecular orbital describes a region of space around two or more atoms inside which electrons are likely to be found.
Mathematical principles tell us that when orbitals combine, the number of orbitals before the combination takes place must equal the number of new orbitals that result from the combination – orbitals don’t just disappear! We saw this previously when we discussed hybrid orbitals: one s and three p orbitals make four sp3 hybrids. When two atomic 1s orbitals combine in the formation of H2, the result is two sigma (σ) orbitals.
Molecular orbitals for H2
According to MO theory, one sigma orbital is lower in energy than either of the two isolated atomic 1s orbitals –this lower sigma orbital is referred to as a bonding molecular orbital. The second, 'sigma star' orbital is higher in energy than the two atomic 1s orbitals, and is referred to as an antibonding molecular orbital.
The bonding sigma orbital, which holds both electrons in the ground state of the molecule, is egg-shaped, encompassing the two nuclei, and with the highest likelihood of electrons being in the area between the two nuclei. The high-energy, antibonding sigma* orbital can be visualized as a pair of droplets, with areas of higher electron density near each nucleus and a ‘node’, (area of zero electron density) midway between the two nuclei.
Remember that we are thinking here about electron behavior as wave behavior. When two separate waves combine, they can do so with constructive interference, where the two amplitudes build up and reinforce one another, or destructive interference, where the two amplitudes cancel one another out. Bonding MOs are the consequence of constructive interference between two atomic orbitals, which results in an attractive interaction and an increase in electron density between the nuclei. Antibonding MO’s are the consequence of destructive interference which results in a repulsive interaction and a region of zero electron density between the nuclei (in other words, a node).
Following the same aufbau ('building up') principle you learned in General Chemistry for writing out electron configurations, we place the two electrons in the H2 molecule in the lowest energy molecular orbital, which is the (bonding) sigma orbital. The bonding (attracting) MO is full, and the antibonding (repulsing) MO is empty.
MO theory and conjugated pi bonds
The advantage of using MO theory to understand bonding in organic molecules becomes more apparent when we think about pi bonds. Let’s first consider the pi bond in ethene from an MO theory standpoint (in this example we will be disregarding the s bonds in the molecule, and thinking only about the π bond). We start with two atomic orbitals: one unhybridized 2p orbital from each carbon. Each contains a single electron. In MO theory, the two atomic combine mathematically to form two pi molecular orbitals, one a low-energy pi bonding orbital and one a high-energy pi* antibonding orbital.
Molecular orbitals for ethene (ethylene)
In the bonding pi orbital, the two shaded lobes of the p orbitals interact constructively with each other, as do the two unshaded lobes (remember, the arbitrary shading choice represents mathematical (+) and (-) signs for the mathematical wavefunction describing the orbital). There is increased electron density between the two carbon nuclei in the molecular orbital - it is a bonding interaction.
In the higher-energy antibonding pi* orbital, the shaded lobe of one p orbital interacts destructively with the unshaded lobe of the second p orbital, leading to a node between the two nuclei and overall repulsion between the carbon nuclei.
Again using the 'building up' principle, we place the two electrons in the lower-energy, bonding pi molecular orbital. The antibonding pi* orbital remains empty.
Next, we'll consider the 1,3-butadiene molecule. From valence orbital theory alone we might expect that the C2-C3 bond in this molecule, because it is a sigma bond, would be able to rotate freely.
Experimentally, however, it is observed that there is a significant barrier to rotation about the C2-C3 bond, and that the entire molecule is planar. In addition, the C2-C3 bond is 148 pm long, shorter than a typical carbon-carbon single bond (about 154 pm), though longer than a typical double bond (about 134 pm).
Molecular orbital theory accounts for these observations with the concept of delocalized pi bonds. In this picture, the four 2p atomic orbitals combine mathematically to form four pi molecular orbitals of increasing energy. Two of these - the bonding pi orbitals - are lower in energy than the p atomic orbitals from which they are formed, while two - the antibonding pi* orbitals - are higher in energy.
The lowest energy molecular orbital, pi1, has only constructive interaction and zero nodes. Higher in energy, but still lower than the isolated p orbitals, the pi2 orbital has one node but two constructive interactions - thus it is still a bonding orbital overall. Looking at the two antibonding orbitals, pi3* has two nodes and one constructive interaction, while pi4* has three nodes and zero constructive interactions.
By the aufbau principle, the four electrons from the isolated 2pz atomic orbitals are placed in the bonding pi1 and pi2 MO’s. Because pi1 includes constructive interaction between C2 and C3, there is a degree, in the 1,3-butadiene molecule, of pi-bonding interaction between these two carbons, which accounts for its shorter length and the barrier to rotation. The valence bond picture of 1,3-butadiene shows the two pi bonds as being isolated from one another, with each pair of pi electrons ‘stuck’ in its own pi bond. However, molecular orbital theory predicts (accurately) that the four pi electrons are to some extent delocalized, or ‘spread out’, over the whole pi system.
space-filling view
1,3-butadiene is the simplest example of a system of conjugated pi bonds. To be considered conjugated, two or more pi bonds must be separated by only one single bond – in other words, there cannot be an intervening sp3-hybridized carbon, because this would break up the overlapping system of parallel p orbitals. In the compound below, for example, the C1-C2 and C3-C4 double bonds are conjugated, while the C6-C7 double bond is isolated from the other two pi bonds by sp3-hybridized C5.
A very important concept to keep in mind is that there is an inherent thermodynamic stability associated with conjugation. This stability can be measured experimentally by comparing the heat of hydrogenation of two different dienes. (Hydrogenation is a reaction type that we will learn much more about in chapter 15: essentially, it is the process of adding a hydrogen molecule - two protons and two electrons - to a p bond). When the two conjugated double bonds of 1,3-pentadiene are 'hydrogenated' to produce pentane, about 225 kJ is released per mole of pentane formed. Compare that to the approximately 250 kJ/mol released when the two isolated double bonds in 1,4-pentadiene are hydrogenated, also forming pentane.
The conjugated diene is lower in energy: in other words, it is more stable. In general, conjugated pi bonds are more stable than isolated pi bonds.
Conjugated pi systems can involve oxygen and nitrogen atoms as well as carbon. In the metabolism of fat molecules, some of the key reactions involve alkenes that are conjugated to carbonyl groups.
In chapter 4, we will see that MO theory is very useful in explaining why organic molecules that contain extended systems of conjugated pi bonds often have distinctive colors. beta-carotene, the compound responsible for the orange color of carrots, has an extended system of 11 conjugated pi bonds.
Exercise: Identify all conjugated and isolated double bonds in the structures below. For each conjugated pi system, specify the number of overlapping p orbitals, and how many pi electrons are shared among them.
Exercise: Identify all isolated and conjugated pi bonds in lycopene, the red-colored compound in tomatoes. How many pi electrons are contained in the conjugated pi system?
Solutions to exercises
Aromaticity - The Ultimate Conjugated System
Molecular orbital theory is especially helpful in explaining the unique properties of aromatic compounds such as benzene:
Although benzene is most often drawn with three double bonds and three single bonds, in fact all of the carbon-carbon bonds iare exactly the same length (138 pm). In addition, the pi bonds in benzene are significantly less reactive than 'normal' pi bonds, either isolated or conjugated. Something about the structure of benzene makes its pi bonding arrangement especially stable. This ‘something’ has a name: it is called ‘aromaticity’.
What exactly is this ‘aromatic’ property that makes the pi bonds in benzene so stable? In large part, the answer to this question lies in the fact that benzene is a cyclic molecule in which all of the ring atoms are sp2-hybridized. This allows the pi electrons to be delocalized in molecular orbitals that extend all the way around the ring, above and below the plane. For this to happen, of course, the ring must be planar – otherwise the p orbitals couldn’t overlap properly. Benzene is indeed known to be a flat molecule.
Do all cyclic molecules with alternating single and double bonds have this same aromatic stability? The answer, in fact, is ‘no’. The eight-membered cyclooctatetraene ring shown below is not flat, and its π bonds react like 'normal' alkenes.
Clearly it takes something more to be aromatic, and this can best be explained with molecular orbital theory. Let’s look at an energy diagram of the pi molecular orbitals in benzene.
Quantum mechanical calculations tell us that the six pi molecular orbitals in benzene, formed from six atomic p orbitals, occupy four separate energy levels. pi1 and pi6* have unique energy levels, while the pi2 - pi3 and pi4*- pi5* pairs are degenerate, meaning they are at the same energy level. When we use the aufbau principle to fill up these orbitals with the six pi electrons in benzene, we see that the bonding orbitals are completely filled, and the antibonding orbitals are empty. This gives us a good clue to the source of the special stability of benzene: a full set of bonding MO’s is similar in many ways to the ‘full shell’ of electrons in the atomic orbitals of the stable noble gases helium, neon, and argon.
Now, let’s do the same thing for cyclooctatetraene, which we have already learned is not aromatic.
The result of molecular orbital calculations tells us that the lowest and highest energy MOs (pi1 and pi8*) have unique energy levels, while the other six form degenerate pairs. Notice that pi4 and pi5 are at the same energy level as the isolated 2pz atomic orbitals: these are therefore neither bonding nor antibonding, rather they are referred to as nonbonding MOs. Filling up the MOs with the eight pi electrons in the molecule, we find that the last two electrons are unpaired and fall into the two degenerate nonbonding orbitals. Because we don't have a perfect filled shell of bonding MOs, our molecule is not aromatic. As a consequence, each of the double bonds in cyclooctatetraene acts more like an isolated double bond.
For now, the important learning objective is to recognize conjugated pi bonds systems and understand the benzene is exceptionally stable exhibiting a property called aromaticity. Aromaticity and chemistry of aromatic compounds is relatively complex and is discussed in greater detail in subsequent chapters of this text.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/02%3A_Structure_and_Properties_of_Organic_Molecules/2.04%3A_2.4_Conjugated_Pi_Bond_Systems.txt |
Learning Objective
• identify the orbitals occupied by lone pair electrons
Valence Bond and Molecular Orbital Theories
The table below summarizes the main points of the two complementary bonding theories. Both theories provide different, useful ways of describing molecular structure. We will use both theories and often blend them to analyze and predict chemical structure and reactivity. The bonding theories are reviewed in greater detail in the next two sections.
Comparison of Bonding Theories
Valence Bond Theory Molecular Orbital Theory
considers bonds as localized between one pair of atoms considers electrons delocalized throughout the entire molecule
creates bonds from overlap of atomic orbitals (s, p, d…) and hybrid orbitals (sp, sp2, sp3…) combines atomic orbitals to form molecular orbitals (σ, σ*, π, π*)
forms σ or π bonds creates bonding and antibonding interactions based on which orbitals are filled
predicts molecular shape based on the number of regions of electron density predicts the arrangement of electrons in molecules
needs multiple structures to describe resonance
Orbitals of Lone Pair Electrons
There are situations in which we will want to integrate molecular orbital and valence bond theories. Identifying the orbitals of lone pair electrons is one situation. Hybridized orbitals create sigma bonds and hold lone pairs. The sigma bonds create the "framework" that holds all the atoms together as a molecule or ion. Un-hybridized p orbitals create pi bonds perpendicular to this sigma framework. In the future, we will learn that some lone pair electrons on heteroatoms of rings can occupy p orbitals to create aromaticity. Stay tuned for upcoming attractions. For the first ten chapters of this text, we will only focus on non-aromatic compounds.
To identify the orbitals of the lone pair electrons in non-aromatic compounds, we can follow a two-step approach.
Step 1: Add any missing lone pair electrons to the heteroatoms (atoms other than carbon and hydrogen).
Step 2: Determine the hybridization of any atoms with lone pairs (heteroatoms). Lone pairs occupy the hybridized orbitals.
Example
To identify the orbitals of the lone pair electrons in the compound below, we will follow the approach above.
Step 1: Add lone pairs.
Step 2: Determine the hybridization of any atom with lone pairs.
The lone pairs on each heteroatom occupy the indicated hybridized orbital.
NOTE: These guidelines only apply for non-aromatic compounds. There can be exceptions to these guidelines for some heterocyclic aromatic compounds. These exceptions are fully explained in a later chapter of this text.
Exercise
1. For the compound below:
a) add the lone pair electrons
b) label the hybridization and electron geometry for all non-hydrogen atoms
c) specify the hybridization of the orbital for each lone pair
d) What is the chemical formula of this compound?
Answer
1.
2.06: Bond Rotation
Learning Objective
• distinguish between bonds that can rotate and those that cannot
Sigma Bonds can Rotate
We learned in section 2.1 that single bonds in organic molecules are free to rotate, due to the 'end-to-end' (sigma) nature of their orbital overlap. Consider the carbon-oxygen bond in ethanol, for example: with a 180o rotation about this bond, the shape of the molecule would look quite different:
For ethane, rotation about the carbon-carbon sigma bond results in many different possible three-dimensional arrangements of the atoms.
Pi Bonds are rigid
Pi bonds are created from overlapping p orbitals. The lobes of the p orbitals prevent the atoms sharing pi bonds from rotating as shown in the diagram below.
Double and Triple Bonds cannot Rotate
The pi bonds in double and triple bonds prevent these bonds from rotating. This rigidity has an effect on the physical structure of compounds and can influence chemical reactivity. For now, we want to build the habit of looking at static drawings and diagrams of organic compounds and visualizing their dynamic nature.
For ethene, there is no rotation about the carbon-carbon double bond because of the pi bond.
Exercise
1. Label the selected bonds in the compound below as "Rotates" or "Rigid."
Answer
1. Arrows for (a) and (c) are pointing to single bonds that can rotate. Arrow (b) is pointing to a double bond that is rigid because of the pi bond. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/02%3A_Structure_and_Properties_of_Organic_Molecules/2.05%3A_Lone_Pair_Electrons_and_Bonding_Theories.txt |
Learning Objective
• recognize the relationships between constitutional (structural) isomers, conformational isomers, and geometric isomers
Isomers
Isomers always have the same chemical formula. When the chemical formulae are different, then the compounds are completely different. Important information can be gained from the chemical formulas when comparing compounds.
Structural (constitutional) isomers have the same molecular formula but a different bonding arrangement among the atoms. Stereoisomers have identical molecular formulas and arrangements of atoms. They differ from each other only in the spatial orientation of groups in the molecule. For organic chemistry, there are several types of stereoisomers: enantiomers, diasteriomers, geometric isomers, and conformers. These stereoisomers will be introduced and explained throughout several chapters.
Structural (Constitutional) Isomers
Because carbon forms four bonds, there can be multiple ways to form molecules that follow the octet rule. Even with only four carbon atoms, there are two possible structures for the carbon backbone. The carbon atoms can be bonded to make a four carbon chain (butane) or there can be a one carbon branch from a three carbon chain (2-methylpropane). Butane and 2-methylpropane are structural isomers because they both have the chemical formula C4H10.
Identical vs Conformer
The rotation about single bonds creates dynamic molecules. When drawing and discussing molecules, it is important to be aware that our drawing are static while the molecule themselves are rotating. Although there are seven sigma bonds in the ethane molecule, rotation about the six carbon-hydrogen bonds does not result in any change in the shape of the molecule because the hydrogen atoms are essentially spherical. Rotation about the carbon-carbon bond, however, results in many different possible molecular conformations. Conformers are the simplest example of stereoisomerism.
Identical compounds are the same compound shown with ALL atoms in the same spatial orientation.
Conformers are the same compound shown with different rotations about single bonds.
In the example below, we can compare two identical structures for ethane with two conformers of ethane.
Geometric Isomers - an example of stereoisomerism
The rigidity of the pi bonds in double bonds can create geometric isomerism. Without rotation, there are two different orientations possible across the carbon-carbon double bond (C=C). The rigidity of the double bond creates a line of reference for spatial orientation. The prefixes cis and trans are used to distinguish between geometric isomers. The cis-stereoisomer has both non-hydrogen atoms on the same side of the double bond. Whereas, the trans-stereoisomer has the non-hydrogen atoms across the double bond. In the same way, we cross the ocean an a trans-Atlantic journey. This small difference may seem insignificant, but geometric isomers are different chemical compounds with different physical properties as shown in the example below.
For now, it is important to distinguish between structural differences and spatial differences when comparing compounds. In the future, we will look more closely at isomerism.
Example
Let's look ate the bond-line structures below and determine the relationships between the following pairs of compounds: identical, conformers, structural isomers, geometric isomers, or completely different compounds.
The first important step (that is often skipped) is to determine the chemical formula of each compound. If the chemical formulas are different, then the compounds are completely different and there is NO isomeric relationship. If the chemical formulas are the same, then we identify the difference between to the compounds to determine their relatioship. If there are structural differences in the bonding patterns, then the compounds are constitutional (structural isomers). If the compounds have the same structural connections, but the spatial orientations are different, then the compounds are stereoisomers. For now, the possible stereoisomers are conformers showing the same compound with different carbon-carbon single bond rotations or geometric isomers of compounds with different orientations at the carbon-carbon double bonds.
Applying the logic above to our example, we determine the following.
Exercise
1. What is the relationship between the following pairs of compounds: identical, conformers, structural isomers, geometric isomers, or completely different compounds?
Answer
1.
a) geometric isomers
b) conformers
c) structural isomers
d) completely different compounds
e) identical | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/02%3A_Structure_and_Properties_of_Organic_Molecules/2.07%3A_Isomerism_Introduction.txt |
Learning Objective
• classify hydrocarbons as saturated or unsaturated
• classify hydrocarbons as alkanes, alkenes, alkynes, cycloalkanes, or aromatics (arenes)
• apply the homologous series to organic molecules with 1-10 carbons
Hydrocarbon Classifications
Hydrocarbons are organic compounds that contain only carbon and hydrogen. The inherent ability of hydrocarbons to bond to themselves is known as catenation, and allows hydrocarbon to form more complex molecules, such as cyclohexane and benzene. Catenation comes from the fact that the bond character between carbon atoms is entirely non-polar.
The four general classes of hydrocarbons are: alkanes, alkenes, alkynes and arenes. Aromatic compounds derive their names from the fact that many of these compounds in the early days of discovery were grouped because they were oils with fragrant odors. The classifications for hydrocarbons are summarized below.
Saturated hydrocarbons (alkanes) are the simplest of the hydrocarbon species. They are composed entirely of single bonds and are saturated with hydrogen. The general formula for saturated hydrocarbons is CnH2n+2 (assuming non-cyclic structures). Saturated hydrocarbons are the basis of petroleum fuels and are found as either linear or branched species.The simplest alkanes have their C atoms bonded in a straight chain; these are called normal alkanes. They are named according to the number of carbon atoms in the chain. The smallest alkane is methane:
Unsaturated hydrocarbons have double and/or triple bonds between carbon atoms. Those with double bond are called alkenes and have the general formula CnH2n (assuming non-cyclic structures). Those containing triple bonds are called alkynes and have general formula CnH2n-2. The smallest alkene—ethene—has two C atoms and is also known by its common name ethylene and the smallest alkyne is ethyne, also known as acetylene.
Cycloalkanes are hydrocarbons containing one or more carbon rings to which hydrogen atoms are attached. The prefix "cyclo" is added to the name to communicate the ring structure. The general formula for a saturated hydrocarbon containing one ring is CnH2n.
Aromatic hydrocarbons, also known as arenes, are hydrocarbons that have at least one aromatic ring. Aromatic compounds contain the benzene unit. Benzene itself is composed of six C atoms in a ring, with alternating single and double C–C bonds:
For most compounds, information beyond the chemical formula will be needed to elucidate their structure. However, the ratio of C:H in a chemical formula can provide insights into the chemical structure.
For example, let's look at some of the possible structures and chemical formulas for hydrocarbons containing six carbon atoms.
The saturated alkane has the highest ratio of hydrogen to carbon. The unsaturated alkene and the six membered alkane ring share the same chemical formula. It is important to remember this relationship. The unsaturated alkyne has a lower ratio of hydrogen to carbon than alkenes with a second pi bond. Benzene rings have the lowest hydrogen ratio to carbon at 1:1.
Exercise
1. Classify the following compounds are saturated or unsaturated. For unsaturated hydrocarbons, refine the classification by indicating whether the compound is an alkene, alkyne, or arene.
Answer
1.
The number of carbons continuously bonded together is an important structural feature and is described using the Homologous Series. In first year organic chemistry, the first ten names of the Homologous Series are usually all that need to be memorized. Of course, your professor will set the standard. Most of the prefixes are familiar from the Greek prefixes for binary covalent compounds. It is the prefixes for the first four carbon chain lengths that may be unfamiliar. Interestingly, three of these hydrocarbons frequently appear in every day life. Methane gas is a primary component of flatulence and is the ingredient that ignites when farts are lit - don't try this at home. Propane and butane are gases at room temperature. They are stored under pressure to create the liquid state. Propane is the fuel for bbqs, while butane is used in lighters. The suffix "ane" is used to distinguish between the longest continuous carbon chain, while the shorter carbon branches (substituents) are indicated with "yl" as the suffix.
Exercise
2. Complete the table below.
Condensed Structural Formula Chemical Name
propane
C6H6
CH3CH2CH2CH2CH2CH3
Answer
2.
Condensed Structural Formula Chemical Name
CH3CH2CH3 propane
C6H6 benzene
CH3CH2CH2CH2CH2CH3 hexane
• Wikipedia | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/02%3A_Structure_and_Properties_of_Organic_Molecules/2.08%3A_Hydrocarbons_and_the_Homologous_Series.txt |
Learning Objective
• recognize and classify the common functional groups of organic chemistry (alkanes, alkenes, alkynes, alkyl halides, alcohols, amines, ethers, aldehydes, ketones, carboxylic acids, esters, and amides
Functional groups in organic compounds
Functional groups are structural units within organic compounds that are defined by specific bonding arrangements between specific atoms. The structure of capsaicin, the compound responsible for the heat in peppers, incorporates several functional groups, labeled in the figure below and explained throughout this section.
As we progress in our study of organic chemistry, it will become extremely important to be able to quickly recognize the most common functional groups, because they are the key structural elements that define how organic molecules react. For now, we will only worry about drawing and recognizing each functional group, as depicted by Lewis and line structures. Much of the remainder of your study of organic chemistry will be taken up with learning about how the different functional groups behave in organic reactions.
The 'default' in organic chemistry (essentially, the lack of any functional groups) is given the term alkane, characterized by single bonds between carbon and carbon, or between carbon and hydrogen. Methane, CH4, is the natural gas you may burn in your furnace. Octane, C8H18, is a component of gasoline.
Alkanes
Alkenes (sometimes called olefins) have carbon-carbon double bonds, and alkynes have carbon-carbon triple bonds. Ethene, the simplest alkene example, is a gas that serves as a cellular signal in fruits to stimulate ripening. (If you want bananas to ripen quickly, put them in a paper bag along with an apple - the apple emits ethene gas, setting off the ripening process in the bananas). Ethyne, commonly called acetylene, is used as a fuel in welding blow torches.
Alkenes and alkynes
Alkenes have trigonal planar electron geometry while alkynes have linear geometry. Furthermore, many alkenes can take two geometric forms: cis or trans. The cis and trans forms of a given alkene are different molecules with different physical properties there is a very high energy barrier to rotation about a double bond. In the example below, the difference between cis and trans alkenes is readily apparent.
Alkanes, alkenes, and alkynes are all classified as hydrocarbons, because they are composed solely of carbon and hydrogen atoms. Alkanes are said to be saturated hydrocarbons, because the carbons are bonded to the maximum possible number of hydrogens - in other words, they are saturated with hydrogen atoms. The double and triple-bonded carbons in alkenes and alkynes have fewer hydrogen atoms bonded to them - they are thus referred to as unsaturated hydrocarbons. As we will see in chapter 15, hydrogen can be added to double and triple bonds, in a type of reaction called 'hydrogenation'.
The aromatic group is exemplified by benzene (which used to be a commonly used solvent on the organic lab, but which was shown to be carcinogenic), and naphthalene, a compound with a distinctive 'mothball' smell. Aromatic groups are planar (flat) ring structures, and are widespread in nature. We will learn more about the structure and reactions of aromatic groups in chapters 2 and 14.
Aromatics
When the carbon of an alkane is bonded to one or more halogens, the group is referred to as a alkyl halide or haloalkane. Chloroform is a useful solvent in the laboratory, and was one of the earlier anesthetic drugs used in surgery. Chlorodifluoromethane was used as a refrigerant and in aerosol sprays until the late twentieth century, but its use was discontinued after it was found to have harmful effects on the ozone layer. Bromoethane is a simple alkyl halide often used in organic synthesis. Alkyl halides groups are quite rare in biomolecules.
Haloalkanes
In the alcohol functional group, a carbon is single-bonded to an OH group (the OH group, by itself, is referred to as a hydroxyl). Except for methanol, all alcohols can be classified as primary, secondary, or tertiary. In a primary alcohol, the carbon bonded to the OH group is also bonded to only one other carbon. In a secondary alcohol and tertiary alcohol, the carbon is bonded to two or three other carbons, respectively. When the hydroxyl group is directly attached to an aromatic ring, the resulting group is called a phenol. The sulfur analog of an alcohol is called a thiol (from the Greek thio, for sulfur).
Alcohols, phenols, and thiols
Note that the definition of a phenol states that the hydroxyl oxygen must be directly attached to one of the carbons of the aromatic ring. The compound below, therefore, is not a phenol - it is a primary alcohol.
The distinction is important, because as we will see later, there is a significant difference in the reactivity of alcohols and phenols.
The deprotonated forms of alcohols, phenols, and thiols are called alkoxides, phenolates, and thiolates, respectively. A protonated alcohol is an oxonium ion.
In an ether functional group, a central oxygen is bonded to two carbons. Below is the structure of diethyl ether, a common laboratory solvent and also one of the first compounds to be used as an anesthetic during operations. The sulfur analog of an ether is called a thioether or sulfide.
Ethers and sulfides
Phosphate and its derivative functional groups are ubiquitous in biomolecules. Phosphate linked to a single organic group is called a phosphate ester; when it has two links to organic groups it is called a phosphate diester. A linkage between two phosphates creates a phosphate anhydride.
Organic phosphates
There are a number of functional groups that contain a carbon-oxygen double bond, which is commonly referred to as a carbonyl. Ketones and aldehydes are two closely related carbonyl-based functional groups that react in very similar ways. In a ketone, the carbon atom of a carbonyl is bonded to two other carbons. In an aldehyde, the carbonyl carbon is bonded on one side to a hydrogen, and on the other side to a carbon. The exception to this definition is formaldehyde, in which the carbonyl carbon has bonds to two hydrogens.
A group with a carbon-nitrogen double bond is called an imine, or sometimes a Schiff base (in this book we will use the term 'imine'). The chemistry of aldehydes, ketones, and imines will be covered in chapter 10.
Aldehydes, ketones, and imines
When a carbonyl carbon is bonded on one side to a carbon (or hydrogen) and on the other side to an oxygen, nitrogen, or sulfur, the functional group is considered to be one of the ‘carboxylic acid derivatives’, a designation that describes a set of related functional groups. The eponymous member of this family is the carboxylic acid functional group, in which the carbonyl is bonded to a hydroxyl group. The conjugate base of a carboxylic acid is a carboxylate. Other derivatives are carboxylic esters (usually just called 'esters'), thioesters, amides, acyl phosphates, acid chlorides, and acid anhydrides. With the exception of acid chlorides and acid anhydrides, the carboxylic acid derivatives are very common in biological molecules and/or metabolic pathways, and their structure and reactivity will be discussed in detail in chapter 11.
Carboxylic acid derivatives
A single compound often contains several functional groups, particularly in biological organic chemistry. The six-carbon sugar molecules glucose and fructose, for example, contain aldehyde and ketone groups, respectively, and both contain five alcohol groups (a compound with several alcohol groups is often referred to as a ‘polyol’).
The hormone testosterone, the amino acid phenylalanine, and the glycolysis metabolite dihydroxyacetone phosphate all contain multiple functional groups, as labeled below.
While not in any way a complete list, this section has covered most of the important functional groups that we will encounter in biological organic chemistry.
Exercise:
1. Identify the functional groups (other than alkanes) in the following organic compounds. State whether alcohols and amines are primary, secondary, or tertiary.
Solution
1.
Amines are characterized by nitrogen atoms with single bonds to hydrogen and carbon. Just as there are primary, secondary, and tertiary alcohols, there are primary, secondary, and tertiary amines. Ammonia is a special case with no carbon atoms.
One of the most important properties of amines is that they are basic, and are readily protonated to form ammonium cations. In the case where a nitrogen has four bonds to carbon (which is somewhat unusual in biomolecules), it is called a quaternary ammonium ion.
Amines
Note: Do not be confused by how the terms 'primary', 'secondary', and 'tertiary' are applied to alcohols and amines - the definitions are different. In alcohols, what matters is how many other carbons the alcohol carbon is bonded to, while in amines, what matters is how many carbons the nitrogen is bonded to.
Finally, a nitrile group is characterized by a carbon triple-bonded to a nitrogen.
Nitriles
Exercise
2. Draw one example of each compound that includes the specified structural features. Be sure to designate the location of all non-zero formal charges. All atoms should have complete octets (phosphorus may exceed the octet rule). There are many possible correct answers for these, so be sure to check your structures with your instructor or tutor.
a) a compound with molecular formula C6H11NO that includes alkene, secondary amine, and primary alcohol functional groups
b) an ion with molecular formula C3H5O6P 2- that includes aldehyde, secondary alcohol, and phosphate functional groups.
c) A compound with molecular formula C6H9NO that has an amide functional group, and does not have an alkene group.
Answer
2. | textbooks/chem/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/02%3A_Structure_and_Properties_of_Organic_Molecules/2.09%3A_Organic_Functional_Groups.txt |
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